GATE EE 2015 Solved Paper

GATE ELECTRICAL ENGINEERING

Solved Paper ( 2014-1996 )

RK Kanodia Ashish Murolia

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NODIA & COMPANY

GATE Electrical Engineering Solved Paper (2014 - 1996 ) RK Kanodia & Ashish Murolia

Copyright © By NODIA & COMPANY Information contained in this book has been obtained by author, from sources believes to be reliable. However, neither NODIA & COMPANY nor its author guarantee the accuracy or completeness of any information herein, and NODIA & COMPANY nor its author shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that NODIA & COMPANY and its author are supplying information but are not attempting to render engineering or other professional services.

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SYLLABUS GENERAL ABILITY Verbal Ability : English grammar, sentence completion, verbal analogies, word groups, instructions, critical reasoning and verbal deduction. Numerical Ability : Numerical computation, numerical estimation, numerical reasoning and data interpretation.

ENGINEERING MATHEMATICS Linear Algebra: Matrix Algebra, Systems of linear equations, Eigen values and eigen vectors. Calculus: Mean value theorems, Theorems of integral calculus, Evaluation of definite and improper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series. Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gauss and Green’s theorems. Differential equations: First order equation (linear and nonlinear), Higher order linear differential equations with constant coefficients, Method of variation of parameters, Cauchy’s and Euler’s equations, Initial and boundary value problems, Partial Differential Equations and variable separable method. Complex variables: Analytic functions, Cauchy’s integral theorem and integral formula, Taylor’s and Laurent’ series, Residue theorem, solution integrals. Probability and Statistics: Sampling theorems, Conditional probability, Mean, median, mode and standard deviation, Random variables, Discrete and continuous distributions, Poisson,Normal and Binomial distribution, Correlation and regression analysis. Numerical Methods: Solutions of non-linear algebraic equations, single and multi-step methods for differential equations. Transform Theory: Fourier transform,Laplace transform, Z-transform.

ELECTRICAL ENGINEERING Electric Circuits and Fields: Network graph, KCL, KVL, node and mesh analysis, transient response of dc and ac networks; sinusoidal steady-state analysis, resonance, basic filter concepts; ideal current and voltage sources, Thevenin’s, Norton’s and Superposition and Maximum Power Transfer theorems, two-port networks, three phase circuits; Gauss Theorem, electric field and potential due to point, line, plane and spherical charge distributions; Ampere’s and Biot-Savart’s laws; inductance; dielectrics; capacitance.

Signals and Systems: Representation of continuous and discrete-time signals; shifting and scaling operations; linear, time-invariant and causal systems; Fourier series representation of continuous periodic signals; sampling theorem; Fourier, Laplace and Z transforms. Electrical Machines: Single phase transformer – equivalent circuit, phasor diagram, tests, regulation and efficiency; three phase transformers – connections, parallel operation; autotransformer; energy conversion principles; DC machines – types, windings, generator characteristics, armature reaction and commutation, starting and speed control of motors; three phase induction motors – principles, types, performance characteristics, starting and speed control; single phase induction motors; synchronous machines – performance, regulation and parallel operation of generators, motor starting, characteristics and applications; servo and stepper motors. Power Systems: Basic power generation concepts; transmission line models and performance; cable performance, insulation; corona and radio interference; distribution systems; per-unit quantities; bus impedance and admittance matrices; load flow; voltage control; power factor correction; economic operation; symmetrical components; fault analysis; principles of overcurrent, differential and distance protection; solid state relays and digital protection; circuit breakers; system stability concepts, swing curves and equal area criterion; HVDC transmission and FACTS concepts. Control Systems: Principles of feedback; transfer function; block diagrams; steady-state errors; Routh and Niquist techniques; Bode plots; root loci; lag, lead and lead-lag compensation; state space model; state transition matrix, controllability and observability. Electrical and Electronic Measurements: Bridges and potentiometers; PMMC, moving iron, dynamometer and induction type instruments; measurement of voltage, current, power, energy and power factor; instrument transformers; digital voltmeters and multimeters; phase, time and frequency measurement; Q-meters; oscilloscopes; potentiometric recorders; error analysis. Analog and Digital Electronics: Characteristics of diodes, BJT, FET; amplifiers – biasing, equivalent circuit and frequency response; oscillators and feedback amplifiers; operational amplifiers – characteristics and applications; simple active filters; VCOs and timers; combinational and sequential logic circuits; multiplexer; Schmitt trigger; multi-vibrators; sample and hold circuits; A/D and D/A converters; 8-bit microprocessor basics, architecture, programming and interfacing. Power Electronics and Drives: Semiconductor power diodes, transistors, thyristors, triacs, GTOs, MOSFETs and IGBTs – static characteristics and principles of operation; triggering circuits; phase control rectifiers; bridge converters – fully controlled and half controlled; principles of choppers and inverters; basis concepts of adjustable speed dc and ac drives.

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PREFACE This book doesn’t make promise but provides complete satisfaction to the readers. The market scenario is confusing and readers don’t find the optimum quality books. This book provides complete set of problems appeared in competition exams as well as fresh set of problems. The book is categorized into units which are then sub-divided into chapters and the concepts of the problems are addressed in the relevant chapters. The aim of the book is to avoid the unnecessary elaboration and highlights only those concepts and techniques which are absolutely necessary. Again time is a critical factor both from the point of view of preparation duration and time taken for solving each problem in the examination. So the problems solving methods is the books are those which take the least distance to the solution. But however to make a comment that this book is absolute for GATE preparation will be an inappropriate one. The theory for the preparation of the examination should be followed from the standard books. But for a wide collection of problems, for a variety of problems and the efficient way of solving them, what one needs to go needs to go through is there in there in the book. Each unit (e.g. Networks) is subdivided into average seven number of chapters on an average each of which contains 40 problems which are selected so as to avoid unnecessary redundancy and highly needed completeness. I shall appreciate and greatly acknowledge the comments and suggestion from the users of this book.

R. K. Kanodia Ashish Murolia

CONTENTS SP 1

Electric Circuit and Fields

SP 1 - 83

SP 2

Signals and Systems

SP 84 - 124

SP 3

Electrical Machines

SP 125 - 200

SP 4

Power Systems

SP 201 - 277

SP 5

Control Systems

SP 278 - 343

SP 6

Electrical & Electronics Measurement SP 344 - 389

SP 7

Analog Electronics

SP 390 - 451

SP 8

Digital Electronics

SP 452 - 493

SP 8

Power Electronics

SP 494 - 551

SP 9

Engineering Mathematics

SP 552 - 595

SP 9

General Aptitude

SP 596 - 623

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For online test series visit

www.gatehelp.com Chapter 1

Electric Circuits and Fields

Page 1

CHAPTER 1 ELECTRIC CIRCUITS AND FIELDS

YEAR 2014 EE01

ONE MARK

EE SP 1.1

The three circuit elements shown in the figure are part of an electric circuit. The total power absorbed by the three circuit elements in watts is _____.

EE SP 1.2

C 0 is the capacitance of a parallel plate capacitor with air as dielectric (as in figure (a)). If, half of the entire gap as shown in figure (b) is filled with a dielectric of permittivity e r , the expression for the modified capacitance is

nodia (A) C 0 ^1 + e r h 2 (C) C 0 e r 2

EE SP 1.3

(B) ^C 0 + e r h (D) C 0 ^1 + e r h

A combination of 1 mF capacitor with an initial voltage vc ^0 h =- 2 V in series with a 100 W resistor is connected to a 20 mA ideal dc current source by operating both switches at t = 0 s as shown. Which of the following graphs shown in the options approximates the voltage vs across the current source over the next few seconds ?

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EE SP 1.4


Chapter 1

nodia The undesirable property of an electrical insulating material is (A) high dielectric strength (B) high relative permittivity (C) high thermal conductivity (D) high insulation resistivity YEAR 2014 EE01

EE SP 1.5

EE SP 1.6

TWO MARKS.

In the figure, the value of resistor R is ^25 + I/2h ohms , where I is the current in amperes. The current I is _____.

The following four vector fields are given in Cartesian co-ordinate system. The vector field which does not satisfy the property of magnetic flux density is (A) y2 ax + z2 ay + x2 az (B) z2 ax + x2 ay + y2 az (C) x2 ax + y2 ay + z2 az (D) y2 z2 ax + x2 z2 ay + x2 y2 az YEAR 2014 EE02

EE SP 1.7

ONE MARK

Two identical coupled inductors are connected in series. The measured inductances for the two possible series connections are 380 mH and 240 mH . Their mutual inductance in mH is _____.


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www.gatehelp.com Chapter 1 EE SP 1.8


Page 3

The switch SW shown in the circuit is kept at position ‘1’ for a long duration. At t = 0+ , the switch is moved to position ‘2’. Assuming Vo2 > Vo1 , the voltage vc ^ t h across the capacitor is

(A) vc ^ t h =- Vo2 ^1 - e-t/2RC h - Vo1 (B) vc ^ t h = Vo2 ^1 - e-t/2RC h + Vo1

(C) vc ^ t h =-^Vo2 + Vo1h^1 - e-t/2RC h - Vo1 (D) vc ^ t h = ^Vo2 - Vo1h^1 - e-t/2RC h + Vo1

EE SP 1.9

nodia

A parallel plate capacitor consisting two dielectric materials is shown in the figure. The middle dielectric slab is placed symmetrically with respect to the plates.

If the potential difference between one of the plates and the nearest surface of dielectric interface is 2 Volts, then the ratio e 1 : e 2 is (A) 1 : 4 (B) 2 : 3 (C) 3 : 2

(D) 4 : 1

YEAR 2014 EE02 EE SP 1.10

TWO MARKS

The voltage across the capacitor, as shown in the figure, is expressed as vc ^ t h = A1 sin ^w 1 t - q 1h + A2 sin ^w 2 t - q 2h

The values of A1 and A2 respectively, are (A) 2.0 and 1.98 (B) 2.0 and 4.20 (C) 2.5 and 3.50 (D) 5.0 and 6.40


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Chapter 1

The magnitude of magnetic flux density ^B h at a point having normal distance d meters from an infinitely extended wire carrying current of I A is 2mpdI (in SI units). An infinitely extended wire is laid along the x -axis and is carrying current of 4 A in the + ve x direction. Another infinitely extended wire is laid along the y axis and is carrying 2 A current in the + ve y direction. m 0 is permeability of free space. Assume i , j , k to be unit vectors along x , y and z axes respectively. 0

nodia

Assuming right handed coordinate system, magnetic field intensity, H at coordinate ^2, 1, 0h will be (A) 3 k weber/m2 (B) 4 i A/m 2p 3p (C) 3 k A/m (D) 0 A/m 2p YEAR 2014 EE03

EE SP 1.12

EE SP 1.13

ONE MARK.

Let d : ^ f v h = x2 y + y2 z + z2 x , where f and v are scalar and vector fields respectively. If v = yi + zj + xk , then v : df is (A) x2 y + y2 z + z2 x (B) 2xy + 2yz + 2zx (C) x + y + z (D) 0 The line A to neutral voltage is 10 15c V for a balanced three phase star connected load with phase sequence ABC . The voltage of line B with respect to line C is given by (A) 10 3 105c V (B) 10 105c V (C) 10 3 - 75c V

EE SP 1.14

(D) - 10 3 90c V

The driving point impedance Z ^s h for the circuit shown below is

4 2 (A) s +3 3s + 1 s + 2s 2 (C) 4 s +2 1 s +s +1

4 2 (B) s +22s + 4 s +2 3 (D) 4 s +2 1 s +s +1


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A non-ideal voltage source Vs has an internal impedance of Zs . If a purely resistive load is to be chosen that maximizes the power transferred to the load, its value must be (A) 0 (B) real part of Zs (C) magnitude of Zs

(D) complex conjugate of Zs


EE SP 1.17

EE SP 1.18

Page 5

TWO MARKS

The power delivered by the current source, in the figure, is ____.

nodia

A perfectly conducting metal plate is placed in x -y plane in a right handed coordinate system. A charge of + 32pe 0 2 coulombs is placed at coordinate ^0, 0, 2h. e 0 is the permittivity of free space. Assume i , j , k to be unit vectors along x , y and z axes respectively. At the coordinate ^ 2 , 2 , 0h, the electric field vector E (Newtons/Coulomb) will be

(A) 2 2 k

(B) - 2k

(C) 2k

(D) - 2 2 k

A series RLC circuit is observed at two frequencies. At w 1 = 1 krad/s , we note that source voltage V1 = 100 0c V results in a current I1 = 0.03 31c A . At w 2 = 2 krad/s , the source voltage V2 = 100 0c V results in a current I2 = 2 0c A . The closest values for R, L, C out of the following options are (A) R = 50 W ; L = 25 mH ; C = 10 mF ; (B) R = 50 W ; L = 10 mH ; C = 25 mF ; (C) R = 50 W ; L = 50 mH ; C = 5 mF ; (D) R = 50 W ; L = 5 mH ; C = 50 mF ;


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Chapter 1

The Norton’s equivalent source in amperes as seen into terminals X and Y is ____.

nodia YEAR 2013

EE SP 1.20

ONE MARK

Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k , k > 0 , the elements of the corresponding star equivalent will be scaled by a factor of

(A) k2

(B) k

(C) 1/k

(D)

k

EE SP 1.21

The flux density at a point in space is given by Bv = 4xavx + 2kyavy + 8avz Wb/m2 . The value of constant k must be equal to (A) - 2 (B) - 0.5 (C) + 0.5 (D) + 2

EE SP 1.22

A single-phase load is supplied by a single-phase voltage source. If the current flowing from the load to the source is 10+ - 150cA and if the voltage at the load terminal is 100+60cV , then the (A) load absorbs real power and delivers reactive power (B) load absorbs real power and absorbs reactive power (C) load delivers real power and delivers reactive power (D) load delivers real power and absorbs reactive power

EE SP 1.23

A source vs ^ t h = V cos 100pt has an internal impedance of ^4 + j3h W . If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in W should be (B) 4 (A) 3 (C) 5 (D) 7


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EE SP 1.24


The transfer function

(A) 0.5s + 1 s+1 (C) s + 2 s+1

V2 ^s h of the circuit shown below is V1 ^s h

(B) 3s + 6 s+2 (D) s + 1 s+2

nodia YEAR 2013

EE SP 1.25

EE SP 1.26

TWO MARKS

A dielectric slab with 500 mm # 500 mm cross-section is 0.4 m long. The slab is subjected to a uniform electric field of Ev = 6avx + 8avy kV/mm . The relative permittivity of the dielectric material is equal to 2. The value of constant e0 is 8.85 # 10-12 F/m . The energy stored in the dielectric in Joules is (B) 8.85 # 10-5 (A) 8.85 # 10-11 (D) 885 (C) 88.5

Three capacitors C1 , C2 and C 3 whose values are 10 mF , 5 mF , and 2 mF respectively, have breakdown voltages of 10 V, 5 V and 2 V respectively. For the interconnection shown below, the maximum safe voltage in Volts that can be applied across the combination, and the corresponding total charge in mC stored in the effective capacitance across the terminals are respectively,

(A) 2.8 and 36 (C) 2.8 and 32 EE SP 1.27

Page 7

(B) 7 and 119 (D) 7 and 80

In the circuit shown below, if the source voltage VS = 100+53.13c V then the Thevenin’s equivalent voltage in Volts as seen by the load resistance RL is


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(A) 100+90c (C) 800+90c

(B) 800+0c (D) 100+60c

YEAR 2012 EE SP 1.28

ONE MARK

The impedance looking into nodes 1 and 2 in the given circuit is

nodia (A) 50 W (C) 5 kW

EE SP 1.29

(B) 100 W (D) 10.1 kW

In the circuit shown below, the current through the inductor is

(B) - 1 A 1+j

2 A 1+j (C) 1 A 1+j (A)

EE SP 1.30

EE SP 1.31

Chapter 1

(D) 0 A

(s2 + 9) (s + 2) (s + 1) (s + 3) (s + 4) is excited by sin (wt). The steady-state output of the system is zero at (A) w = 1 rad/s (B) w = 2 rad/s (C) w = 3 rad/s (D) w = 4 rad/s A system with transfer function G (s) =

The average power delivered to an impedance (4 - j3) W by a current 5 cos (100pt + 100) A is (A) 44.2 W (B) 50 W (C) 62.5 W (D) 125 W


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Page 9

In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current i (t) for all t is

(A) zero

(B) a step function

(C) an exponentially decaying function (D) an impulse function YEAR 2012

EE SP 1.33

nodia If VA - VB = 6 V then VC - VD is

(A) - 5 V (C) 3 V

EE SP 1.34

TWO MARKS

(B) 2 V (D) 6 V

Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is

(A) 0.8 W (C) 2 W

(B) 1.4 W (D) 2.8 W

Common Data for Questions 35 and 36 : With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed : (i) 1 W connected at port B draws a current of 3 A (ii) 2.5 W connected at port B draws a current of 2 A


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Chapter 1

EE SP 1.35

With 10 V dc connected at port A, the current drawn by 7 W connected at port B is (A) 3/7 A (B) 5/7 A (C) 1 A (D) 9/7 A

EE SP 1.36

For the same network, with 6 V dc connected at port A, 1 W connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is (A) 6 V (B) 7 V (C) 8 V

(D) 9 V

Statement for Linked Answer Questions 37 and 38 :

nodia

In the circuit shown, the three voltmeter readings are V1 = 220 V, V2 = 122 V, V3 = 136 V .

EE SP 1.37

EE SP 1.38

The power factor of the load is (A) 0.45 (C) 0.55

(B) 0.50 (D) 0.60

If RL = 5 W , the approximate power consumption in the load is (A) 700 W (B) 750 W (C) 800 W (D) 850 W YEAR 2011

EE SP 1.39

ONE MARK

The r.m.s value of the current i (t) in the circuit shown below is

(A) 1 A 2 (C) 1 A

(B) 1 A 2 (D) 2 A


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The voltage applied to a circuit is 100 2 cos (100pt) volts and the circuit draws a current of 10 2 sin (100pt + p/4) amperes. Taking the voltage as the reference phasor, the phasor representation of the current in amperes is (A) 10 2 - p/4 (B) 10 - p/4 (C) 10 + p/4

EE SP 1.41

(D) 10 2 + p/4

In the circuit given below, the value of R required for the transfer of maximum power to the load having a resistance of 3 W is

nodia (B) 3 W (D) infinity

(A) zero

(C) 6 W

YEAR 2011

EE SP 1.42

EE SP 1.43

Page 11

TWO MARKS

A lossy capacitor Cx , rated for operation at 5 kV, 50 Hz is represented by an equivalent circuit with an ideal capacitor C p in parallel with a resistor R p . The value C p is found to be 0.102 μF and value of R p = 1.25 MW . Then the power loss and tan d of the lossy capacitor operating at the rated voltage, respectively, are (A) 10 W and 0.0002 (B) 10 W and 0.0025 (C) 20 W and 0.025 (D) 20 W and 0.04 A capacitor is made with a polymeric dielectric having an er of 2.26 and a dielectric breakdown strength of 50 kV/cm. The permittivity of free space is 8.85 pF/m. If the rectangular plates of the capacitor have a width of 20 cm and a length of 40 cm, then the maximum electric charge in the capacitor is (A) 2 μC (B) 4 μC (C) 8 μC (D) 10 μC

Common Data For Q. 44 and 45 The input voltage given to a converter is vi = 100 2 sin (100pt) V The current drawn by the converter is ii = 10 2 sin (100pt - p/3) + 5 2 sin (300pt + p/4) + 2 2 sin (500pt - p/6) A EE SP 1.44

The input power factor of the converter is (A) 0.31 (B) 0.44 (C) 0.5 (D) 0.71

EE SP 1.45

The active power drawn by the converter is (A) 181 W (B) 500 W (C) 707 W (D) 887 W


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Chapter 1

Common Data For Q. 46 and 47 An RLC circuit with relevant data is given below.

EE SP 1.46

The power dissipated in the resistor R is (A) 0.5 W (B) 1 W

nodia (C)

2W (D) 2 W

EE SP 1.47

The current IC in the figure above is (A) - j2 A

(B) - j

(C) + j

(D) + j2A

1 2

A

1 2

A


The switch in the circuit has been closed for a long time. It is opened at t = 0. At t = 0+ , the current through the 1 mF capacitor is

(A) 0 A (C) 1.25 A EE SP 1.49

ONE MARK

(B) 1 A (D) 5 A

As shown in the figure, a 1 W resistance is connected across a source that has a load line v + i = 100 . The current through the resistance is

(A) 25 A (C) 100 A

(B) 50 A (C) 200 A


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TWO MARKS

If the 12 W resistor draws a current of 1 A as shown in the figure, the value of resistance R is

(A) 4 W (C) 8 W EE SP 1.51

(B) 6 W (D) 18 W

nodia

The two-port network P shown in the figure has ports 1 and 2, denoted by terminals (a,b) and (c,d) respectively. It has an impedance matrix Z with parameters denoted by Zij . A 1 W resistor is connected in series with the network at port 1 as shown in the figure. The impedance matrix of the modified two-port network (shown as a dashed box ) is

Z11 + 1 (A) e Z21 Z11 + 1 (C) e Z21

Z12 + 1 Z22 + 1o Z12 Z22 o

Z11 + 1 Z12 (B) e Z21 Z22 + 1o Z11 + 1 Z12 (D) e Z21 + 1 Z22 o


ONE MARK

The current through the 2 kW resistance in the circuit shown is

(A) 0 mA (C) 2 mA EE SP 1.53

Page 13

(B) 1 mA (D) 6 mA

How many 200 W/220 V incandescent lamps connected in series would consume the same total power as a single 100 W/220 V incandescent lamp ? (A) not possible (B) 4 (C) 3 (D) 2


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EE SP 1.55

TWO MARKS

In the figure shown, all elements used are ideal. For time t < 0, S1 remained closed and S2 open. At t = 0, S1 is opened and S2 is closed. If the voltage Vc2 across the capacitor C2 at t = 0 is zero, the voltage across the capacitor combination at t = 0+ will be

(A) 1 V

(B) 2 V

(C) 1.5 V

(D) 3 V

nodia

The equivalent capacitance of the input loop of the circuit shown is

(A) 2 mF (C) 200 mF EE SP 1.56

Chapter 1

(B) 100 mF (D) 4 mF

For the circuit shown, find out the current flowing through the 2 W resistance. Also identify the changes to be made to double the current through the 2 W resistance.

(A) (5 A; PutVS = 30 V) (C) (5 A; Put IS = 10 A)

(B) (2 A; PutVS = 8 V) (D) (7 A; Put IS = 12 A)

Statement for Linked Answer Question 57 and 58 :


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Page 15

EE SP 1.57

For the circuit given above, the Thevenin’s resistance across the terminals A and B is (A) 0.5 kW (B) 0.2 kW (C) 1 kW (D) 0.11 kW

EE SP 1.58

For the circuit given above, the Thevenin’s voltage across the terminals A and B is (A) 1.25 V (B) 0.25 V (C) 1 V (D) 0.5 V YEAR 2008

EE SP 1.59

ONE MARK

The number of chords in the graph of the given circuit will be

nodia (A) 3 (C) 5

EE SP 1.60

(B) 4 (D) 6

The Thevenin’s equivalent of a circuit operation at w = 5 rads/s, has Voc = 3.71+ - 15.9% V and Z0 = 2.38 - j0.667 W . At this frequency, the minimal realization of the Thevenin’s impedance will have a (A) resistor and a capacitor and an inductor (B) resistor and a capacitor (C) resistor and an inductor (D) capacitor and an inductor YEAR 2008

EE SP 1.61

The time constant for the given circuit will be

(A) 1/9 s (C) 4 s EE SP 1.62

TWO MARKS

(B) 1/4 s (D) 9 s

The resonant frequency for the given circuit will be


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EE SP 1.63


(A) 1 rad/s

(B) 2 rad/s

(C) 3 rad/s

(D) 4 rad/s

Chapter 1

Assuming ideal elements in the circuit shown below, the voltage Vab will be

(A) - 3 V (C) 3 V

(B) 0 V

nodia (D) 5 V

Statement for Linked Answer Question 64 and 65

The current i (t) sketched in the figure flows through a initially uncharged 0.3 nF capacitor.

EE SP 1.64

The charge stored in the capacitor at t = 5 ms , will be (A) 8 nC (B) 10 nC (C) 13 nC (D) 16 nC

EE SP 1.65

The capacitor charged upto 5 ms, as per the current profile given in the figure, is connected across an inductor of 0.6 mH. Then the value of voltage across the capacitor after 1 ms will approximately be (A) 18.8 V (B) 23.5 V (C) - 23.5 V (D) - 30.6 V


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EE SP 1.67


Page 17

In the circuit shown in the figure, the value of the current i will be given by

(A) 0.31 A

(B) 1.25 A

(C) 1.75 A

(D) 2.5 A

Two point charges Q1 = 10 mC and Q2 = 20 mC are placed at coordinates (1,1,0) and (- 1, - 1, 0) respectively. The total electric flux passing through a plane z = 20 will be (A) 7.5 mC (B) 13.5 mC (C) 15.0 mC

nodia (D) 22.5 mC

EE SP 1.68

EE SP 1.69

A capacitor consists of two metal plates each 500 # 500 mm2 and spaced 6 mm apart. The space between the metal plates is filled with a glass plate of 4 mm thickness and a layer of paper of 2 mm thickness. The relative primitivities of the glass and paper are 8 and 2 respectively. Neglecting the fringing effect, the capacitance will be (Given that e0 = 8.85 # 10 - 12 F/m ) (A) 983.3 pF (B) 1475 pF (C) 637.7 pF (D) 9956.25 pF

A coil of 300 turns is wound on a non-magnetic core having a mean circumference of 300 mm and a cross-sectional area of 300 mm2. The inductance of the coil corresponding to a magnetizing current of 3 A will be (Given that m0 = 4p # 10 - 7 H/m) (A) 37.68 mH (B) 113.04 mH (C) 3.768 mH (D) 1.1304 mH YEAR 2007

EE SP 1.70

ONE MARK

Divergence of the vector field V (x, y, z) =- (x cos xy + y) it + (y cos xy) jt + (sin z2 + x2 + y2) kt is (A) 2z cos z2 (B) sin xy + 2z cos z2 (C) x sin xy - cos z (D) None of these


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Chapter 1

TWO MARKS

The state equation for the current I1 in the network shown below in terms of the voltage VX and the independent source V , is given by

(A) dI1 =- 1.4VX - 3.75I1 + 5 V 4 dt (B) dI1 = 1.4VX - 3.75I1 - 5 V dt 4 (C) dI1 =- 1.4VX + 3.75I1 + 5 V 4 dt (D) dI1 =- 1.4VX + 3.75I1 - 5 V dt 4

EE SP 1.72

EE SP 1.73

nodia

The R-L-C series circuit shown in figure is supplied from a variable frequency voltage source. The admittance - locus of the R-L-C network at terminals AB for increasing frequency w is

In the circuit shown in figure. Switch SW1 is initially closed and SW2 is open. The inductor L carries a current of 10 A and the capacitor charged to 10 V with polarities as indicated. SW2 is closed at t = 0 and SW1 is opened at t = 0 . The


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Page 19

current through C and the voltage across L at (t = 0+) is

EE SP 1.74

EE SP 1.75

EE SP 1.76

(A) 55 A, 4.5 V

(B) 5.5 A, 45 V

(C) 45 A, 5.5 A

(D) 4.5 A, 55 V

A 3 V DC supply with an internal resistance of 2 W supplies a passive non-linear 2 . The power dissipated in the resistance characterized by the relation VNL = INL non linear resistance is (A) 1.0 W (B) 1.5 W (C) 2.5 W (D) 3.0 W

nodia

In the figure given below all phasors are with reference to the potential at point ''O'' . The locus of voltage phasor VYX as R is varied from zero to infinity is shown by

The matrix A given below in the node incidence matrix of a network. The columns correspond to branches of the network while the rows correspond to nodes. Let V = [V1V2 .....V6]T denote the vector of branch voltages while I = [i1 i2 .....i6]T that of branch currents. The vector E = [e1 e2 e3 e4]T denotes the vector of node voltages relative to a common ground.


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Chapter 1

R1 1 1 0 0 0 V W S S 0 -1 0 -1 1 0 W S- 1 0 0 0 - 1 - 1W W S S 0 0 -1 1 0 1 W X T Which of the following statement is true ? (A) The equations V1 - V2 + V3 = 0,V3 + V4 - V5 = 0 are KVL equations for the network for some loops (B) The equations V1 - V3 - V6 = 0,V4 + V5 - V6 = 0 are KVL equations for the network for some loops (C) E = AV (D) AV = 0 are KVI equations for the network EE SP 1.77

nodia

A solid sphere made of insulating material has a radius R and has a total charge Q distributed uniformly in its volume. What is the magnitude of the electric field intensity, E , at a distance r (0 < r < R) inside the sphere ? 1 Qr 4pe0 R3 Q (C) 1 2 4pe0 r

3 Qr 4pe0 R3 QR (D) 1 4pe0 r3

(A)

(B)


An inductor designed with 400 turns coil wound on an iron core of 16 cm2 cross sectional area and with a cut of an air gap length of 1 mm. The coil is connected to a 230 V, 50 Hz ac supply. Neglect coil resistance, core loss, iron reluctance and leakage inductance, (m0 = 4p # 10 - 7 H/M)

EE SP 1.78

EE SP 1.79

The current in the inductor is (A) 18.08 A (C) 4.56 A

(B) 9.04 A (D) 2.28 A

The average force on the core to reduce the air gap will be (A) 832.29 N (B) 1666.22 N (C) 3332.47 N (D) 6664.84 N YEAR 2006

EE SP 1.80

ONE MARK

In the figure the current source is 1+0 A, R = 1 W , the impedances are ZC =- j W and ZL = 2j W . The Thevenin equivalent looking into the circuit across X-Y is


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(A)

2 +0 V, (1 + 2j) W

(C) 2+45% V, (1 + j) W

Page 21

(B) 2+45% V, (1 - 2j) W (D)

2 +45% V, (1 + j) W


TWO MARKS

The parameters of the circuit shown in the figure are Ri = 1 MW R0 = 10 W, A = 106 V/V If vi = 1 mV , the output voltage, input impedance and output impedance respectively are

nodia (A) 1 V, 3, 10 W (C) 1 V, 0, 3

EE SP 1.82

(B) 1 V, 0, 10 W (D) 10 V, 3, 10 W

In the circuit shown in the figure, the current source I = 1 A , the voltage source V = 5 V, R1 = R2 = R3 = 1 W, L1 = L2 = L3 = 1 H, C1 = C2 = 1 F

The currents (in A) through R3 and through the voltage source V respectively will be (A) 1, 4 (B) 5, 1 (C) 5, 2 (D) 5, 4 EE SP 1.83

The parameter type and the matrix representation of the relevant two port parameters that describe the circuit shown are

0 0 (A) z parameters, = G 0 0 0 0 (C) h parameters, = G 0 0

1 0 (B) h parameters, = G 0 1 1 0 (D) z parameters, = G 0 1


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Chapter 1

The circuit shown in the figure is energized by a sinusoidal voltage source V1 at a frequency which causes resonance with a current of I .

The phasor diagram which is applicable to this circuit is

nodia EE SP 1.85

An ideal capacitor is charged to a voltage V0 and connected at t = 0 across an ideal inductor L. (The circuit now consists of a capacitor and inductor alone). If 1 , the voltage across the capacitor at time t > 0 is given by we let w0 = LC (A) V0 (B) V0 cos (w0 t) (C) V0 sin (w0 t)

(D) V0 e - w t cos (w0 t) 0

EE SP 1.86

An energy meter connected to an immersion heater (resistive) operating on an AC 230 V, 50 Hz, AC single phase source reads 2.3 units (kWh) in 1 hour. The heater is removed from the supply and now connected to a 400 V peak square wave source of 150 Hz. The power in kW dissipated by the heater will be (A) 3.478 (B) 1.739 (C) 1.540 (D) 0.870

EE SP 1.87

Which of the following statement holds for the divergence of electric and magnetic flux densities ? (A) Both are zero (B) These are zero for static densities but non zero for time varying densities. (C) It is zero for the electric flux density (D) It is zero for the magnetic flux density


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ONE MARK

In the figure given below the value of R is

(A) 2.5 W (C) 7.5 W EE SP 1.89

Page 23

(B) 5.0 W (D) 10.0 W

nodia The RMS value of the voltage u (t)= 3 + 4 cos (3t) is (A) 17 V (B) 5 V (C) 7 V (D) (3 + 2 2 ) V

EE SP 1.90

EE SP 1.91

For the two port network shown in the figure the Z -matrix is given by

Z1 Z1 + Z2 (A) = Z1 + Z2 Z2 G

Z1 Z1 (B) = Z1 + Z2 Z2 G

Z1 Z2 (C) = Z2 Z1 + Z2 G

Z1 Z1 (D) = Z1 Z1 + Z2 G

In the figure given, for the initial capacitor voltage is zero. The switch is closed at t = 0 . The final steady-state voltage across the capacitor is

(A) 20 V (C) 5 V EE SP 1.92

(B) 10 V (D) 0 V

If Ev is the electric intensity, 4 (4 # Ev) is equal to (A) Ev (B) Ev (C) null vector (D) Zero


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Chapter 1

YEAR 2005

TWO MARKS

Statement for Linked Answer Question 93 and 94 A coil of inductance 10 H and resistance 40 W is connected as shown in the figure. After the switch S has been in contact with point 1 for a very long time, it is moved to point 2 at, t = 0 . EE SP 1.93

If, at t = 0+ , the voltage across the coil is 120 V, the value of resistance R is

nodia (A) 0 W (C) 40 W

EE SP 1.94

EE SP 1.95

EE SP 1.96

(B) 20 W (D) 60 W

For the value as obtained in (a), the time taken for 95% of the stored energy to be dissipated is close to (A) 0.10 sec (B) 0.15 sec (C) 0.50 sec (D) 1.0 sec The RL circuit of the figure is fed from a constant magnitude, variable frequency sinusoidal voltage source Vin . At 100 Hz, the Rand L elements each have a voltage drop mRMS .If the frequency of the source is changed to 50 Hz, then new voltage drop across R is

(A)

5 8

uRMS

(B)

2 3

uRMS

(C)

8 5

uRMS

(D)

3 2

uRMS

For the three-phase circuit shown in the figure the ratio of the currents IR: IY : IB is given by


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(A) 1 : 1 : (C) 1 : 1 : 0 EE SP 1.97

3

Page 25

(B) 1 : 1 : 2 (D) 1 : 1 : 3/2

The circuit shown in the figure is in steady state, when the switch is closed at t = 0 .Assuming that the inductance is ideal, the current through the inductor at t = 0+ equals

nodia (A) 0 A

(B) 0.5 A (C) 1 A (D) 2 A

EE SP 1.98

The charge distribution in a metal-dielectric-semiconductor specimen is shown in the figure. The negative charge density decreases linearly in the semiconductor as shown. The electric field distribution is as shown in


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Chapter 1

In the given figure, the Thevenin’s equivalent pair (voltage, impedance), as seen at the terminals P-Q, is given by

(A) (2 V, 5 W) (B) (2 V, 7.5 W) (C) (4 V, 5 W)

nodia (D) (4 V, 7.5 W)

YEAR 2004

EE SP 1.100

The value of Z in figure which is most appropriate to cause parallel resonance at 500 Hz is

(A) 125.00 mH (C) 2.0 mF EE SP 1.101

(B) 304.20 mF (D) 0.05 mF

A parallel plate capacitor is shown in figure. It is made two square metal plates of 400 mm side. The 14 mm space between the plates is filled with two layers of dielectrics of er = 4 , 6 mm thick and er = 2 , 8 mm thick. Neglecting fringing of fields at the edge the capacitance is

(A) 1298 pF (C) 354 pF EE SP 1.102

ONE MARK

(B) 944 pF (D) 257 pF

The inductance of a long solenoid of length 1000 mm wound uniformly with 3000 turns on a cylindrical paper tube of 60 mm diameter is (A) 3.2 mH (B) 3.2 mH (D) 3.2 H (C) 32.0 mH


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YEAR 2004 EE SP 1.103

EE SP 1.104

(A) 12 V

(B) 24 V

(C) 30 V

(D) 44 V

In figure, Ra , Rb and Rc are 20 W, 20 W and 10 W respectively. The resistances R1 , R2 and R 3 in W of an equivalent star-connection are

nodia (B) 5, 2.5, 5 (D) 2.5, 5, 2.5

In figure, the admittance values of the elements in Siemens are YR = 0.5 + j0, YL = 0 - j1.5, YC = 0 + j0.3 respectively. The value of I as a phasor when the voltage E across the elements is 10+0% V

(A) 1.5 + j0.5 (C) 0.5 + j1.8

EE SP 1.106

TWO MARKS

In figure, the value of the source voltage is

(A) 2.5, 5, 5 (C) 5, 5, 2.5

EE SP 1.105

Page 27

(B) 5 - j18 (D) 5 - j12

In figure, the value of resistance R in W is

(A) 10 (C) 30

(B) 20 (D) 40


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EE SP 1.108

EE SP 1.109


In figure, the capacitor initially has a charge of 10 Coulomb. The current in the circuit one second after the switch S is closed will be

(A) 14.7 A

(B) 18.5 A

(C) 40.0 A

(D) 50.0 A

The rms value of the current in a wire which carries a d.c. current of 10 A and a sinusoidal alternating current of peak value 20 A is (A) 10 A (B) 14.14 A (C) 15 A (D) 17.32 A

nodia

0.9 0.2 The Z-matrix of a 2-port network as given by = 0.2 0.6G The element Y22 of the corresponding Y-matrix of the same network is given by (A) 1.2 (B) 0.4 (C) - 0.4 (D) 1.8 YEAR 2003

EE SP 1.110

ONE MARK

Figure Shows the waveform of the current passing through an inductor of resistance 1 W and inductance 2 H. The energy absorbed by the inductor in the first four seconds is

(A) 144 J (C) 132 J EE SP 1.111

Chapter 1

(B) 98 J (D) 168 J

A segment of a circuit is shown in figure vR = 5V, vc = 4 sin 2t .The voltage vL is


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(A) 3 - 8 cos 2t (C) 16 sin 2t EE SP 1.112

(B) 60+30%

nodia (C) 70+30%

(D) 34.4+65%

Two conductors are carrying forward and return current of +I and - I as shown in figure. The magnetic field intensity H at point P is

(A) I Y pd (C) I Y 2pd EE SP 1.114

(B) 32 sin 2t (D) 16 cos 2t

In the figure, Z1 = 10+ - 60%, Z2 = 10+60%, Z3 = 50+53.13% . Thevenin impedance seen form X-Y is

(A) 56.66+45%

EE SP 1.113

Page 29

(B) I X pd (D) I X 2pd

Two infinite strips of width w m in x -direction as shown in figure, are carrying forward and return currents of +I and - I in the z - direction. The strips are separated by distance of x m. The inductance per unit length of the configuration is measured to be L H/m. If the distance of separation between the strips in snow reduced to x/2 m, the inductance per unit length of the configuration is

(A) 2L H/m (C) L/2 H/m

(B) L/4 H/m (D) 4L H/m


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YEAR 2003 EE SP 1.115

EE SP 1.116

Chapter 1

TWO MARKS

In the circuit of figure, the magnitudes of VL and VC are twice that of VR . Given that f = 50 Hz , the inductance of the coil is

(A) 2.14 mH

(B) 5.30 H

(C) 31.8 mH

(D) 1.32 H

nodia In figure, the potential difference between points P and Q is

(A) 12 V (B) 10 V (C) - 6 V (D) 8 V EE SP 1.117

Two ac sources feed a common variable resistive load as shown in figure. Under the maximum power transfer condition, the power absorbed by the load resistance RL is

(A) 2200 W (B) 1250 W (C) 1000 W (D) 625 W


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Page 31

In figure, the value of R is

(A) 10 W (B) 18 W (C) 24 W

nodia (D) 12 W

EE SP 1.119

In the circuit shown in figure, the switch S is closed at time (t = 0). The voltage across the inductance at t = 0+ , is

(A) 2 V (B) 4 V (C) - 6 V (D) 8 V EE SP 1.120

The h-parameters for a two-port network are defined by E1 h11 h12 I1 = I G = =h h G =E G 2 21 22 2 For the two-port network shown in figure, the value of h12 is given by

(A) 0.125 (B) 0.167 (C) 0.625 (D) 0.25


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A point charge of +I nC is placed in a space with permittivity of 8.85 # 10 - 12 F/m as shown in figure. The potential difference VPQ between two points P and Q at distance of 40 mm and 20 mm respectively fr0m the point charge is

(A) 0.22 kV (C) - 2.24 kV EE SP 1.122

EE SP 1.123

Chapter 1

(B) - 225 V (D) 15 V

A parallel plate capacitor has an electrode area of 100 mm2, with spacing of 0.1 mm between the electrodes. The dielectric between the plates is air with a permittivity of 8.85 # 10 - 12 F/m. The charge on the capacitor is 100 V. The stored energy in the capacitor is (A) 8.85 pJ (B) 440 pJ (C) 22.1 nJ (D) 44.3 nJ

nodia

A composite parallel plate capacitor is made up of two different dielectric material with different thickness (t1 and t2 ) as shown in figure. The two different dielectric materials are separated by a conducting foil F. The voltage of the conducting foil is

(A) 52 V (C) 67 V

(B) 60 V (D) 33 V ***********


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Page 33

SOLUTION SOL 1.1

Correct answer is 330 W. We redraw the given electric circuit as

nodia

From the given circuit, we have the following observations for the connected batteries: 100 V battery: As the current flows through the battery from positive to negative terminal, so it absorbs power. 80 V battery: As the current flows through the battery from negative to positive terminal, so it delivers power. 15 V battery: As the current flows through the battery negative to positive terminal, so it delivers power. Thus, the net power absorbed by circuit elements is obtained as Pnet = Power absorbed by 100 V battery – [Power emitted by 80 V battery + Power emitted by 15 V battery] = 100 # 10 - 6^80 # 8h + ^15 # 2h@ = 1000 - 640 - 30 = 330 W

SOL 1.2

Correct option is (A). We redraw the given parallel plate capacitors as

(a)

(b)

For the capacitor shown in figure (a), we have C0 = eo A d

...(1)

A = Area of the parallel plate capacitor d = distance between the plates Again, the capacitor shown in figure (b) can be considered as the two capacitors C1 and C2 connected in parallel. So, we have where,


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C1 =

Chapter 1

eo d

A 2

eo e r d Therefore, we get the net capacitance as and

C2 =

A 2

C net = C1 + C2 Ae o /2 Ae o e r /2 Ae o Ae o e r + = + = Ae o ^1 + e r h d d 2d 2d 2d Thus, from equations (1) and (2), we get C net = Co ^1 + e r h 2 =

SOL 1.3

...(2)

nodia Correct answer is (C) We have the given circuit as

In the given circuit, the switches operates as shown by the respective arrows. So, at t = 0 , Switch 1 (switch across Vs ) changes it state from short circuit to open circuit. Switch 2 changes it state from open circuit to short circuit. Therefore, we have the equivalent circuit at t = 0- as (given Vc ^0 h =- 2 V )

So, from the circuit, we get Vs = 0 V t = 0 Again, we consider the circuit for t = 0+ . By s -domain analysis, we have the equivalent voltage across capacitor as I ^s h Vc ^0+h s ^ h V = + S Sc So, we draw the equivalent circuit for given initial voltage across capacitor as -


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Page 35

Therefore, we redraw the complete circuit for t = 0+ as

Applying KVL, we get -2 b 5 l + I + 100I = Vs Sc

nodia - 2 + 20 # 10 s s Thus, we obtain

or

-3

#

1 20 # 10-3 V 100 + # s s s # 1006 3

Vs =- 2 + 20 #2 10 + 2 s s s = 22 # 10 4 s So, Vs ^ t h = 2 # 10 4 t i.e. the plot of source voltage is a straight line passing through origin, as shown below.

SOL 1.4

Correct option is (B). Undesirable property of electrical insulating material is its high relative permittivity ê r h.

SOL 1.5

Correct answer is 10 A. We redraw the given circuit as

Applying KVL in the circuit, we have 300 = IR 300 = I b 25 + I l 2


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Chapter 1 2

300 = 25I + I 2 I 2 + 50I - 600 = 0 I 2 + 60I - 10I - 600 = 0

I Î + 60h - 10 Î + 60h = 0 Î - 10hÎ + 60h = 0 So, I = 10 A , - 60 A Now, for the two obtained values of current, we get R = b 25 + 10 l = 30 2 R = b 25 - 60 l =- 5 2 negative)

(Resistance can not be

nodia Thus, the current through the circuit is

I = 10 A

SOL 1.6

Correct option is (C). According to Maxwell fourth equation

D:B = 0 i.e. the divergence of magnetic field is zero. Now, we check this property for each of the given vectors. Option (A):

So, Option (B): So, Option (C): So,

B = y 2 a x + z 2 ay + x 2 a z D : B = 2 y2 + 2 z2 + 2 x2 = 0 2x 2y 2z B = z 2 a x + x 2 ay + y 2 a z D : B = 2 z2 + 2 x2 + 2 y2 = 0 2x 2y 2z B = x 2 a x + y 2 ay + z 2 a z D : B = 2 x2 + 2 y2 + 2 z2 2x 2y 2z = 2x + 2y + 2z ! 0

Option (D): B = y 2 z 2 a x + x 2 z 2 ay + x 2 y 2 a z D : B = 2 y2 z2 + 2 x2 z2 + 2 x2 y2 = 0 2x 2y 2z Thus, the vector given in option (C) does not satisfy the property of magnetic flux density. So,

SOL 1.7

Correct answer is 35. There are two possible connections (1) Series adding connection


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Page 37

Equivalent inductance Leq = L1 + L2 + 2M 380 = L + L + 2M or L + M = 190 (2) Series opposing connection

...(1)

Equivalent inductance Leq = L1 + L2 - 2M 240 = L + L - 2M or L - M = 120 Substituting eq (2) from eq (1), we get

nodia

...(2)

2L = 70

or

SOL 1.8

L = 35 mH

Correct option is (D). For t < 0 , the switch was at position ‘1’ and circuit is in steady state. In steady state capacitor behaves as open circuit. So

Therefore vC ^0-h =- Vo1 Step 2 : For t > 0 , the switch moves to position 2. Again in steady state capacitor acts as an open circuit.

vC ^3h = Vo2 Time constant t = RTh C RTh is the Thevenin resistance across the capacitor terminal after t = 0 . So t = 2RC Now, at any time t > 0 , capacitor voltage is given by vC ^ t h = vC ^3h + 6vC ^0 h - vC ^3h@e-t/t


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Chapter 1

= Vo2 + 6- Vo1 - Vo2@e-t/2RC = Vo2 + 6- Vo1 - Vo2@e-t/2RC

or,

vC ^ t h = Vo2 ^1 - e- 2RC h - Vo1 e- 2RC t

or,

t

vC ^ t h = Vo2 ^1 - e- 2RC h - Vo1 e- 2RC - Vo1 ^1 - e- 2RC h + Vo1 ^1 - e-t/2RC h t

t

vC ^ t h = ^Vo2 - Vo1h^1 - e- 2RC h + Vo1 t

or, SOL 1.9

t

Correct option is (C).

nodia

As shown in Figure above this arrangement is equivalent to series combination of three capacitors C1 , C2 and C 3 . We have C1 = Ae 1 = 4Ae 1 d d/4 C2 = Ae 2 = 2Ae 2 d d/2 C 3 = Ae 1 = 4Ae 1 d d/4 Now given that voltage across first capacitor is C1 is 2 V, so we write 1 /jwC 1 V =2 VC1 = 1 + 1 + 1 # s j w C 1 j wC 2 j w C 3 C2 C3 10 = 2 C 2 C 3 + C1 C 3 + C1 C 2 #

^Vs = 10 Vh

Substituting values of C1 , C2 and C 3 , we get 8e 1 e 2 # 10 = 2 8e 1 e 2 + 16e 12 + 8e 1 e 2 8e 1 e 2 =1 5 16e 1 e 2 + 16e 12 e2 =1 2e 2 + 2e 1 5 2 e 2 + 2 e 1 = 5e 2 2e 1 = 3e 2 e1 = 3 e2 2 e1 : e 2 = 3 : 2 SOL 1.10

Correct option is (A). Since there are two sources with different frequencies ( w 1 = 10 , w 2 = 5 ), we use


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Page 39

superposition to obtain vC ^ t h. First consider source v ^ t h = 20 sin 10t . Open circuit current source and redrawing circuit in domain with w 1 = 10 rad/ sec .

1 j10 (Using voltage division) VC1 = # 20 1+ 1 j10 1 = 20 1 + j10 # VC1 = 20 - tan-1 10 101 A1 = VC1 = 20 = 1.99 A 101 Now, consider current source i ^ t h = 10 sin 5t . Short circuit voltage source and redrawing circuit in frequency domain with w 2 = 5 rad/ sec .

nodia IC2 =

1 10 0c = j5 10 0c 1 + j5 # 1+ 1 j5

VC2 = IC2 # 1 j5 10 0c VC2 = 1 + j5 VC2 = 10 - tan-5 26 A2 = VC2 = 10 = 1.96 A 26 SOL 1.11



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Chapter 1

H By wire on x -axis = I k 2pd =4# 1 k 1 2p H by wire on y -axis = I ^- k h 2pd = 2 # 1 ^- k h 2 2p H on (2, 1, 0) = 4 k - 1 k 2p 2p H = 3 k A/m 2p SOL 1.12

Correct option is (A). Given the relation

nodia

...(1) d : ^ f v h = x2 y + y2 z + z2 x where f , v are scalar and vector field respectively. From the properties of vector field, we have d : ^ f v h = v : ^df h - f ^d : v h

or Again, we have So,

v : ^df h = d : ^ f v h + f ^d : v h

...(2)

v = yi + zj + xk 2y 2z 2x d:v = + + 2x 2y 2z =0

Therefore, we get

f ^d : v h = 0 Substituting it in equation (2), we get

v : ^df h = d : ^ f v h

Thus, SOL 1.13

v : ^df h = x2 y + y2 z + z2 x

[from equation (1)]

Correct option is (C). We sketch the phasor diagram for three phase star connected load as

Since, line A to neutral voltage is A = 10 15c V


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Page 41

So, for balance 3-phase, we have B = 10 - 105c and C = 10 - 225c Therefore, the voltage of line B with respect to line C is vBC = vB - vC = 10 - 105c - 10 - 225c = 10 3 - 75c SOL 1.14

Correct option is (A). We have the impedance circuit,

nodia

The capacitor and inductor can be replaced with its equivalent reactance (in frequency domain) as shown below.

So, we have the equivalent circuit as

Solving the above circuit, we get the equivalent driving point impedance as Z ^s h = s + 1 || bs + 1 l s s 2 = s + 1 || b s + 1 l s s

or

1 s2 + 1 # s = s+ s 1 + s2 + 1 s s 2 + s 1 = s+ 3 s + 2s 4 2 Z ^s h = s +3 3s + 1 s + 2s


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www.gatehelp.com Page 42 SOL 1.15


Chapter 1

Correct option is (A). According to Gauss’s law, the total electric flux coming out of the concentric spherical surface is equal to the charge enclosed by the surface, i.e.

#

Q = D : ds Since, the electric potential is defined as Q v = 4pe 0 r Q = v # 4pe 0 r = 4pe 0 r

So,

# D : ds = 4pe r

Hence, SOL 1.16

(since, v = 1 volt)

0

Correct answer is 2. To obtain the Norton equivalent along X -Y terminal, we redraw the given circuit as

nodia

In the network, 5 W resistor is short-circuit. So, the equivalent circuit is

Now, writing the KVL equation in two loops, we have 5 - 5I1 - 5 Î1 - Isc h = 0 - 5 Îsc - I1h - 2.5Isc + 2.5 = 0 Solving equation (1),

10I1 - 5Isc = 5 2I1 = Isc + 1

...(1) ...(2)

...(3)

Solving equation (2), or

7.5Isc - 5I1 = 2.5 3Isc - 2I1 = 1


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Page 43

[using equation (3)] 3Isc - Isc - 1 = 1 2Isc = 2 Thus, Isc = 1 A i.e. the Norton equivalent source as seen into terminals X and Y is 1 A.

or or

SOL 1.17

Correct answer is 3 W. To obtain the required unknown, we redraw the given circuit as

nodia Applying KCL at node Vx , we have

Vx - 1 + Vx = 2 1 1

2Vx - 1 = 2 or Vx = 3 2 Therefore, the power delivered by current source is obtained as = Voltage across current source # current through or

current source = 3#2 = 3W 2 SOL 1.18

Correct option is (D). We have the given system of a charge and a perfectly conducting metal plate as shown below.

Due to presence of conducting plane, we can assume an image charge for the system as shown below


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Chapter 1

So, we can observe from the figure that electric field vector along x -y plane is cancelled, i.e. zero. Therefore, the electric field vector acts along negative z -direction and given as " ...(1) E = d E1 + E2 n^- kth 2 2 where E1 and E2 are the field intensity due to the charge Q and - Q at point ( 2 , 2 , 0). So, we obtain Q 32pe 0 2 E1 = 1 2 = 1 4pe 0 r 4pe 0 ( ( 2 ) 2 + ( 2 ) 2 + (2) 2) 2 = 2 Similarly, we get, E2 = 2 Substituting these values in equation (1), we obtain

nodia "

2 + 2 =- 2kt

E =e

SOL 1.19

2 - kt o^ h 2

Correct option is (C). Consider the circuit shown below with non-ideal voltage source Vs , an internal impedance Zs = Rs + jXs , and purely resistive load RL .

Current through the circuit is given by Vs Rs + RL + jXs Vs I = ^Rs + RL h + ^Xs h2 I = or

Therefore, the power transferred to load is P = I 2 RL =

V s2 RL ^^Rs + RL h2 + X s2h

For maximum power, we have


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Page 45

dP = 0 dRL

or

[^Rs + RL h2 + X s2 ] 1 - 2RL ^Rs + RL h =0 2 ^^Rs + RL h2 + X s2h

or

R s2 + R L2 + 2Rs RL + X s2 - 2Rs RL - 2R L2 = 0 R s2 + X s2 = R L2

or

or RL = R s2 + X s2 i.e. to maximise the power transferred to pure resistive load (RL ), its value must be equal to the magnitude of Zs of internal circuit. SOL 1.20

Correct option is (B). Given the current and voltage at different frequencies as At w 1 = 1 krad /s, V1 = 100 0 I1 = 0.03 31c At w 2 = 2 krad /s, V2 = 100 0 I 2 = 2 0c From the given data, it is clear that voltage and current are in phase at w 2 = 2 . So, it satisfies the resonance condition. Therefore, we have R = V = 100 = 50 W 2 I Also, for the resonance circuit, we have wL = 1 wC 1 or 103 # 2 # L = 2 # C # 103 or ...(1) 4 # 106 L = 1 C Again, at w = 1 krad/s, we have cos 31c = R Z R or cos 31c = 2 2 R + b wL - 1 l wC

nodia cos 31c =

or

50 2500 + b103 L -

1 103 C l

...(2)

Solving equations (1) and (2), we get L = 10 mH and C . 25 mF SOL 1.21

Correct option is (B). In the equivalent star connection, the resistance can be given as RC =

Rb Ra Ra Rc Rb Rc ; RB = and RA = Ra + Rb + Rc Ra + Rb + Rc Ra + Rb + Rc

So, if the delta connection components Ra , Rb and Rc are scaled by a factor k then 2 ^k Rb h^k Rc h Rb Rc = k RA =k RAl = kRa + kRb + kRc k Ra + Rb + Rc Hence, it is also scaled by a factor k


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Chapter 1

Correct option is (A). Given, flux density Bv = 4x avx + 2ky avy + 8 avz Since, magnetic flux density is always divergence less. i.e., d$Bv = 0 So, for given vector flux density, we have d$Bv = 4 + 2k + 0 = 0 k =- 2

or, SOL 1.23

Correct option is (B). Consider the voltage source and load shown in figure

nodia We obtain the power delivered by load as

Pdelivered = I L* VL = ^10 + 150ch^10 60ch = 100 210c = 1000 cos 210c + j1000 sin 210c =- 866.025 - j500 As both the reactive and average power (real power) are negative so, power is absorbed by load. i.e., load absorbs real as well as reactive power.

SOL 1.24

SOL 1.25

Correct option is (C). For the purely resistive load, maximum average power is transferred when where RTh + jXTh

2 2 + XTh RL = RTh is the equivalent thevinin (input) impedance of the circuit.

i.e.,

RL =

42 + 32 = 5 W

Correct option is (D). For the given capacitance, C = 100mF in the circuit, we have the reactance. 4 1 = 10 XC = 1 = 6 sc s s # 100 # 10

So,

SOL 1.26

10 4 + 10 4 V2 ^s h s+1 = 4 s 4 = s+2 V1 ^s h 10 + 10 4 + 10 s s

Correct option is (B). Energy density stored in a dielectric medium is obtained as wE = 1 e E 2 J/m2 2 The electric field inside the dielectric will be same to given field in free space


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Page 47

only if the field is tangential to the interface 2 So, wE = 1 2e0 ^ 62 + 82 h # 106 /mm2 2 Therefore, the total stored energy is WE =

#W v

E

dv = e0 100 # 106 /mm2 # ^500 # 500h mm2 # ^0.4h

= e0 # 100 # 106 # 0.4 # 25 # 10 4 = 8.85 # 10-12 # 1013 = 88.5 J SOL 1.27


nodia

Consider that the voltage across the three capacitors C1 , C 2 and C 3 are V1 , V2 and V3 respectively. So, we can write V2 = C 3 ....(1) V3 C2 Since, Voltage is inversely proportional to capacitance Now, given that

C1 = 10 mF ; ^V1hmax = 10V C2 = 5 mF ; ^V2hmax = 5 V C 3 = 2 mF ; ^V3hmax = 2V So, from Eq (1) we have V2 = 2 5 V3 for We obtain,

^V3hmax = 2 V2 = 2 # 2 = 0.8 volt < 5 5

i.e., V2 < ^V2hmax Hence, this is the voltage at C2 . Therefore, V3 = 2 volt V2 = 0.8 volt and V1 = V2 + V3 = 2.8 volt Now, equivalent capacitance across the terminal is Ceq = C 2 C 3 + C1 C2 + C3 = 5 # 2 + 10 = 80 mF 5+2 7 Equivalent voltage is (max. value) Vmax = V1 = 2.8 So, charge stored in the effective capacitance is Q = Ceq Vmax = b 80 l # ^2.8h = 32 mC 7


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Chapter 1

Correct option is (C). For evaluating the equivalent thevenin voltage seen by the load RL , we open the circuit across it (also if it consist dependent source). The equivalent circuit is shown below

As the circuit open across RL so I2 = 0 or, j40I2 = 0 i.e., the dependent source in loop 1 is short circuited. Therefore, ^ j4h Vs VL1 = j4 + 3 j40 100 53.13c VTh = 10 VL1 = j4 + 3 40 90c 100 53.13c = 5 53.13c = 800 90c

nodia SOL 1.29


Applying nodal analysis at top node. V1 + 1 0c V1 + 1 0c = 1 0c + 1 j1 V1 (j 1 + 1) + j 1 + 1 0c = j1 V1 = - 1 1 + j1 Current

1 V1 + 1 0c - 1 + j + 1 = I1 = j1 j1 j = = 1 A (1 + j) j 1 + j


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www.gatehelp.com Chapter 1 SOL 1.30


Page 49

Correct option is (C). (s2 + 9) (s + 2) (s + 1) (s + 3) (s + 4) (- w2 + 9) (jw + 2) G (jw) = (jw + 1) (jw + 3) (jw + 4) The steady state output will be zero if G (s) =

G (jw) = 0 -w 2 + 9 = 0 w = 3 rad/s SOL 1.31

Correct option is (A). We put a test source between terminal 1, 2 to obtain equivalent impedance

nodia ZTh = Vtest Itest

Applying KCL at top right node Vtest + Vtest - 99I = I test b 9 k + 1k 100 Vtest + Vtest - 99I = I test b 10 k 100

...(i)

Ib =- Vtest =-Vtest 9k + 1k 10k

But

Substituting Ib into equation (i), we have Vtest + Vtest + 99Vtest = I test 10 k 100 10 k 100Vtest + Vtest = I test 10 # 103 100 2Vtest = I test 100 Vtest ZTh = Itest = 50 W SOL 1.32

Correct option is (B). In phasor form

Z = 4 - j3 Z = 5 - 36.86cW I = 5 100c A Average power delivered : Pavg. = 1 I 2 Z cos q 2 = 1 # 25 # 5 cos 36.86c = 50 W 2


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Chapter 1

Alternate Method: Z = (4 - j3) W I = 5 cos (100pt + 100) A 2 Pavg = 1 Re $ I Z . 2 = 1 # Re "(5) 2 # (4 - j3), 2 = 1 # 100 = 50 W 2 SOL 1.33

Correct option is (D). The s -domain equivalent circuit is shown as below.

nodia vc (0) /s v (0) = c 1 + 1 1 + 1 C1 C 2 C1 s C 2 s = b C1 C2 l (12 V) C1 + C 2

I (s) =

vC (0) = 12 V

= 12Ceq Taking inverse Laplace transform for the current in time domain, i (t) = 12Ceq d (t) SOL 1.34

(Impulse)

Correct option is (A). In the given circuit,

VA - VB = 6 V So current in the branch, IAB = 6 = 3 A 2 We can see, that the circuit is a one port circuit looking from terminal BD as shown below

For a one port network current entering one terminal, equals the current leaving the second terminal. Thus the outgoing current from A to B will be equal to the incoming current from D to C as shown i.e. IDC = IAB = 3 A


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Page 51

The total current in the resistor 1 W will be I1 = 2 + IDC

(Writing KCL at node D )

= 2+3 = 5A VCD = 1 # (- I1) =- 5 V

So, SOL 1.35

nodia Correct option is (A). We obtain Thevenin equivalent of circuit B .

Thevenin Impedance :

ZTh = R Thevenin Voltage : VTh = 3 0c V Now, circuit becomes as

I1 = 10 - 3 2+R Power transfer from circuit A to B P = (I 12) 2 R + 3I1 Current in the circuit,


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Chapter 1

2 = :10 - 3D R + 3 :10 - 3D = 49R 2 + 21 2+R 2+R (2 + R) (2 + R) 49R + 21 (2 + R) = = 42 + 70R2 2 (2 + R) (2 + R) 2 dP = (2 + R) 70 - (42 + 70R) 2 (2 + R) = 0 dR (2 + R) 4 (2 + R) [(2 + R) 70 - (42 + 70R) 2] = 0

140 + 70R - 84 - 140R = 0 56 = 70R R = 0.8 W SOL 1.36

Correct option is (C). When 10 V is connected at port A the network is

nodia

Now, we obtain Thevenin equivalent for the circuit seen at load terminal, let Thevenin voltage is VTh, 10 V with 10 V applied at port A and Thevenin resistance is RTh .

IL =

VTh,10 V RTh + RL

For RL = 1 W , IL = 3 A 3=

VTh,10 V RTh + 1

...(i)

2=

VTh,10 V RTh + 2.5

...(ii)

For RL = 2.5 W , IL = 2 A

Dividing above two 3 = RTh + 2.5 2 RTh + 1

Substituting RTh

3RTh + 3 = 2RTh + 5 RTh = 2 W into equation (i) VTh,10 V = 3 (2 + 1) = 9 V


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Page 53

Note that it is a non reciprocal two port network. Thevenin voltage seen at port B depends on the voltage connected at port A. Therefore we took subscript VTh,10 V . This is Thevenin voltage only when 10 V source is connected at input port A. If the voltage connected to port A is different, then Thevenin voltage will be different. However, Thevenin’s resistance remains same. Now, the circuit is

nodia For RL = 7 W ,

SOL 1.37

IL =

VTh,10 V = 9 = 1A 2 + RL 2 + 7

Correct option is (B). Now, when 6 V connected at port A let Thevenin voltage seen at port B is VTh,6 V . Here RL = 1 W and IL = 7 A 3

VTh, 6 V = RTh # 7 + 1 # 7 = 2 # 7 + 7 = 7 V 3 3 3 3 This is a linear network, so VTh at port B can be written as VTh = V1 a + b where V1 is the input applied at port A. We have V1 = 10 V , VTh,10 V = 9 V ...(i) 9 = 10a + b When V1 = 6 V , VTh, 6 V = 9 V ...(ii) 7 = 6a + b Solving (i) and (ii) a = 0.5 , b = 4 Thus, with any voltage V1 applied at port A, Thevenin voltage or open circuit voltage at port B will be So, VTh, V = 0.5V1 + 4 For V1 = 8 V 1

VTh,8 V = 0.5 # 8 + 4 = 8 = Voc SOL 1.38

(open circuit voltage)

Correct option is (A). By taking V1, V2 and V3 all are phasor voltages. V1 = V2 + V3 Magnitude of V1, V2 and V3 are given as


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Chapter 1

V1 = 220 V , V2 = 122 V , V3 = 136 V Since voltage across R is in same phase with V1 and the voltage V3 has a phase difference of q with voltage V1 , we write in polar form V1 = V2 0c + V3 q V1 = V2 + V3 cos q + jV3 sin q V1 = (V2 + V3 cos q) + jV3 sin q V1 =

(V2 + V3 cos q) 2 + (V2 sin q) 2

220 =

(122 + 136 cos q) 2 + (136 sin q) 2

Solving, power factor cos q = 0.45 SOL 1.39

Correct option is (B). Voltage across load resistance

nodia Power absorbed in RL

SOL 1.40

VRL = V3 cos q = 136 # 0.45 = 61.2 V 2 (61.2) 2 - 750 W PL = V RL = 5 RL

Correct option is (B). The frequency domains equivalent circuit at w = 1 rad/ sec .

Since the capacitor and inductive reactances are equal in magnitude, the net impedance of that branch will become zero. Equivalent circuit

Current rms value of current SOL 1.41

i (t) = sin t = (1 sin t) A 1W i rms = 1 A 2

Correct option is (D). Voltage in time domain v (t) = 100 2 cos (100pt) Current in time domain


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Page 55

i (t) = 10 2 sin (100pt + p/4) Applying the following trigonometric identity sin (f) = cos (f - 90c) So,

i (t) = 10 2 cos (100pt + p/4 - p/2) = 10 2 cos (100pt - p/4) I = 10 2 - p/4 2

In phasor form, SOL 1.42

Correct option is (A).

nodia Power transferred to the load

2 10 RL l Rth + RL For maximum power transfer Rth , should be minimum. Rth = 6R = 0 6+R

P = I 2 RL = b

R =0 Note: Since load resistance is constant so we choose a minimum value of Rth

SOL 1.43


2 (5 103) 2 Power loss = V rated = # = 20 W Rp 1.25 # 106 For an parallel combination of resistance and capacitor 1 1 = tan d = wC p R p 2p # 50 # 1.25 # 0.102 = 1 = 0.025 40

SOL 1.44

Correct option is (C). Charge

Q = CV = e0 er A V = (e0 er A)V d d

C = e0 er A d

Q = Q max We have e0 = 8.85 # 10 F/cm , er = 2.26 , A = 20 # 40 cm2 -14


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Chapter 1

V = 50 103 kV/cm # d Maximum electrical charge on the capacitor V = V when b d l = 50 kV/cm d max Thus,

Q = 8.85 # 10-14 # 2.26 # 20 # 40 # 50 # 103 = 8 mC

SOL 1.45

Correct option is (C). vi = 100 2 sin (100pt) V Fundamental component of current ii = 10 2 sin (100pt - p/3) A 1

nodia Input power factor

I1 (rms) cos f1 Irms I1 (rms) is rms values of fundamental component of current and Irms is the rms value of converter current. 10 pf = cos p/3 = 0.44 2 10 + 52 + 22 pf =

SOL 1.46

Correct option is (B). Only the fundamental component of current contributes to the mean ac input power. The power due to the harmonic components of current is zero. So, Pin = Vrms I1rms cos f1 = 100 # 10 cos p/3 = 500 W

SOL 1.47

Correct option is (B). Power delivered by the source will be equal to power dissipated by the resistor. P = Vs Is cos p/4 = 1#

SOL 1.48

2 cos p/4 = 1 W

Correct option is (D). IC = Is - I RL = =

2 p/ 4 -

2 - p/ 4

2 $^cos p/4 + j sin p/4h - ^cos p/4 - j sin p/4h.

= 2 2 j sin p/4 = 2j SOL 1.49

Correct option is (B). For t < 0 , the switch was closed for a long time so equivalent circuit is


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Page 57

Voltage across capacitor at t = 0 5 =4 V 4#1 Now switch is opened, so equivalent circuit is vc (0) =

For capacitor at t = 0+ vc (0+) = vc (0) = 4 V v (0+) current in 4 W resistor at t = 0+ , i1 = c =1A 4

nodia so current in capacitor at t = 0+ , ic (0+) = i1 = 1 A

SOL 1.50

Correct option is (B). Thevenin equivalent across 1 X resistor can be obtain as following Open circuit voltage vth = 100 V (i = 0) (vth = 0 ) Short circuit current isc = 100 A So, Rth = vth = 100 = 1 W isc 100 Equivalent circuit is

i = 100 = 50 A 1+1

SOL 1.51

Correct option is (B). The circuit is

Current in R W resistor is i = 2-1 = 1 A Voltage across 12 W resistor is VA = 1 # 12 = 12 V So, i = VA - 6 = 12 - 6 = 6 W 1 R


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Chapter 1


V l1 = Zl11 I l1 + Zl12 I l2 V1 = Z11 I1 + Z12 I2 V l2 = Zl21 I l1 + Zl22 I l2 V2 = Z21 I1 + Z22 I2 Here, I1 = I l1, I2 = I l2 When R = 1 W is connected V l1 = V1 + I l1 # 1 = V1 + I1 V l1 V l1 So, Zl11 Zl12 Similarly for output port V l2 So, Zl21 = Z21 , Zl22 = Z22

nodia = Z11 + 1 = Z12

= Zl21 I l1 + Zl22 I l2 = Zl21 I1 + Zl22 I2

Z11 + 1 Z12 Z => Z21 Z22H

Z-matrix is

SOL 1.53

= Z11 I1 + Z12 I2 + I1 = (Z11 + 1) I1 + Z12 I2


In the bridge R1 R 4 = R 2 R 3 = 1 So it is a balanced bridge I = 0 mA SOL 1.54

Correct option is (D). Resistance of the bulb rated 200 W/220 V is 2 (220) 2 = 242 W R1 = V = 200 P1 Resistance of 100 W/220 V lamp is 2 (220) 2 = 484 W RT = V = 100 P2 To connect in series RT = n # R1 484 = n # 242 n =2


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Page 59

Correct option is (D). For t < 0 , S1 is closed and S2 is opened so the capacitor C1 will charged upto 3 volt. VC1 (0) = 3 Volt Now when switch positions are changed, by applying charge conservation Ceq VC (0+) = C1 VC (0+) + C2 VC (0+) (2 + 1) # 3 = 1 # 3 + 2 # VC (0+) 9 = 3 + 2VC (0+) VC (0+) = 3 Volt 1

1

2

2

2

2

SOL 1.56


nodia Applying KVL in the input loop v1 - i1 (1 + 1) # 103 - 1 (i1 + 49i1) = 0 jwC

Input impedance, Equivalent capacitance,

v1 = 2 # 103 i1 + 1 50i1 jw C 1 Z1 = v1 = 2 # 103 + i1 jw (C/50) 100 nF Ceq = C = = 2 nF 50 50

SOL 1.57

Correct option is (B). Voltage across 2 X resistor, VS = 2 V Current, I2W = VS = 4 = 2 A 2 2 To make the current double we have to take VS = 8 V

SOL 1.58

Correct option is (B). To obtain equivalent Thevenin circuit, put a test source between terminals AB

Applying KCL at super node


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Chapter 1

VP - 5 + VP + VS = I S 2 2 1 VP - 5 + VP + 2VS = 2IS 2VP + 2VS = 2Is + 5 VP + VS = IS + 2.5 & So,

VP - VS = 3VS VP = 4VS 4VS + VS = IS + 2.5

5VS = IS + 2.5 VS = 0.2IS + 0.5 For Thevenin equivalent circuit

nodia VS = IS Rth + Vth By comparing (2) and (3), Thevenin resistance Rth = 0.2 kW

SOL 1.59

SOL 1.60

...(1)

Correct option is (D). From above

...(2)

...(3)

Vth = 0.5 V

Correct option is (A). No. of chords is given as l = b-n+1 b " no. of branches n " no. of nodes l " no. of chords b = 6, n = 4 l = 6 - 4 + 1= 3

SOL 1.61

Correct option is (A). Impedance Zo = 2.38 - j0.667 W Constant term in impedance indicates that there is a resistance in the circuit. Assume that only a resistance and capacitor are in the circuit, phase difference in Thevenin voltage is given as (Due to capacitor) q =- tan- 1 (wCR) j Zo = R wC 1 = 0.667 So, wC and

R = 2.38 W


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Page 61

q =- tan- 1 b 1 # 2.38 l =- 74.34c =[ 15.9c 0.667 given Voc = 3.71+ - 15.9c So, there is an inductor also connected in the circuit SOL 1.62

Correct option is (C). Time constant of the circuit can be calculated by simplifying the circuit as follows

nodia Ceq = 2 F 3

Equivalent Resistance

Req = 3 + 3 = 6 W

Time constant SOL 1.63

t = Req Ceq = 6 # 2 = 4 sec 3

Correct option is (C). Impedance of the circuit is 1 j wC 1 j wC

R 1 - jwCR R = jwL + # 1 - jwCR 1 + w j CR +R R (1 - jwCR) = j wL + 1 + w2 C2 R2 jwL (1 + w2 C2 R2) + R - jwCR2 = 1 + w2 C2 R2 j [wL (1 + w2 C2 R2) - wCR2] R = + 1 + w2 C2 R2 1 + w2 C2 R2 For resonance Im (Z) = 0 So, wL (1 + w2 C2 R2) = wCR2 L = 0.1 H, C = 1 F, R = 1 W So, w # 0.1 [1 + w2 (1) (1)] = w (1) (1) 2 Z = jw L +

1 + w2 = 10 & SOL 1.64

w=

9 = 3 rad/sec

Correct option is (A). Applying KVL in the circuit i = 1 A,

Vab - 2i + 5 = 0 Vab = 2 # 1 - 5 =- 3 Volt


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Chapter 1

Correct option is (C). Charge stored at t = 5 m sec 5

Q =

# i (t) dt

=area under the curve

0

nodia

Q =Area OABCDO =Area (OAD)+Area(AEB)+Area(EBCD) = 1#2#4+1#2#3+3#2 2 2 = 4 + 3 + 6 = 13 nC

SOL 1.66

Correct option is (D). Initial voltage across capacitor Q V0 = o = 13 nC = 43.33 Volt 0.3 nF C When capacitor is connected across an inductor it will give sinusoidal esponse as where At t = 1 m sec ,

SOL 1.67

vc (t) = Vo cos wo t 1 wo = 1 = -9 LC 0.3 # 10 # 0.6 # 10- 3 = 2.35 # 106 rad/sec vc (t) = 43.33 cos (2.35 # 106 # 1 # 10- 6) = 43.33 # (- 0.70) =- 30.44 V

Correct option is (B). Writing node equations at node A and B Va - 5 + Va - 0 = 0 1 1 2Va - 5 = 0 Va = 2.5 V Similarly Vb - 4Vab ++Vb - 0 = 0 3 1 Vb - 4 (Va - Vb) + Vb = 0 3 Vb - 4 (2.5 - Vb) + 3Vb = 0 8Vb - 10 = 0


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Page 63

Vb = 1.25 V i = Vb = 1.25 A 1

Current SOL 1.68

Correct option is ().

SOL 1.69

Correct option is (B). Here two capacitance C1 and C2 are connected in series, so equivalent capacitance is Ceq = C1 C2 C1 + C 2

nodia - 12 -6 # 500 # 10 C1 = e0 er1 A = 8.85 # 10 # 8 # 500 d1 4 # 10- 3

= 442.5 # 10- 11 F - 12 -6 # 500 # 10 C2 = e0 er2 A = 8.85 # 10 # 2 # 500 3 d2 2 # 10

= 221.25 # 10- 11 F - 11 - 11 Ceq = 442.5 # 10 - 11 # 221.25 # 10- 11 442.5 # 10 + 221.25 # 10 = 147.6 # 10- 11 - 1476 pF

SOL 1.70

Correct option is (B). Circumference no. of turns Cross sectional area Inductance of coil

l = 300 mm n = 300 A = 300 mm2 m n2 A L = 0 l 4p # 10- 7 # (300) 2 # 300 # 10- 6 (300 # 10- 3) = 113.04 mH =

SOL 1.71

Correct option is (A). Divergence of a vector field is given as Divergence = 4: V In cartesian coordinates 4 = 2 it + 2 tj + 2 kt 2x 2y 2z So 4: V = 2 6- (x cos xy + y)@ + 2 6(y cos xy)@ + 2 6(sin z2 + x2 + y2)@ 2x 2y 2z =- x (- sin xy) y + y (- sin xy) x + 2z cos z2 = 2z cos z2


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Chapter 1

Correct option is (A). Writing KVL for both the loops V - 3 (I1 + I2) - Vx - 0.5 dI1 = 0 dt V - 3I1 - 3I2 - Vx - 0.5 dI1 = 0 dt

...(1)

In second loop - 5I2 + 0.2Vx + 0.5 dI1 = 0 dt I2 = 0.04Vx + 0.1 dI1 dt

...(2)

Put I2 from eq(2) into eq(2) V - 3I1 - 3 :0.04Vx + 0.1 dI1 D - Vx - 0.5 dI1 = 0 dt dt 0.8 dI1 =- 1.12Vx - 3I1 + V dt dI1 =- 1.4V - 3.75I + 5 V x 1 4 dt

SOL 1.73

nodia Correct option is (A). Impedance of the given network is

Z = R + j b wL - 1 l wC 1 AdmittanceY = 1 = Z R + j b wL - 1 l wC R - j ^wL - w1C h R - j ^wL - w1C h 1 = = # R + j ^wL - w1C h R - j ^wL - w1C h R2 + ^wL - w1C h2 j ^wL - w1C h R = 2 R + ^wL - w1C h2 R2 + ^wL - w1C h2 = Re (Y) + Im (Y) Varying frequency for Re (Y) and Im (Y) we can obtain the admittance-locus.

SOL 1.74

Correct option is (D). At t = 0+ , when switch positions are changed inductor current and capacitor voltage does not change simultaneously So at t = 0+ vc (0+) = vc (0-) = 10 V iL (0+) = iL (0-) = 10 A The equivalent circuit is


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Page 65

Applying KCL vL (0+) vL (0+) - vc (0+) + = iL (0+) = 10 10 10 2vL (0+) - 10 = 100 Voltage across inductor at t = 0+ vL (0+) = 100 + 10 = 55 V 2

nodia So, current in capacitor at t = 0+

vL (0+) - vc (0+) iC (0 ) = 10 = 55 - 10 = 4.5 A 10 +

SOL 1.75

Correct option is (A). The circuit is

Applying KVL 2 = VNL 3 - 2 # I NL 2 2 3 - 2I NL = I NL 2 3I NL = 3 & INL = 1 A VNL = (1) 2 = 1 V

SOL 1.76

Correct option is (B). In the circuit

VX = V+0c


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Chapter 1

Vy - 2V+0c + (Vy) jwC = 0 R Vy (1 + jwCR) = 2V+0c Vy = 2V+0c 1 + jwCR VYX = VX - VY = V -

2V 1 + jwCR

VYX = V - 2V =- V

R " 0,

R " 3, VYX = V - 0 = V So power dissipated in the non-linear resistance P = VNL INL = 1 # 1 = 1 W SOL 1.77

Correct option is (C). In node incidence matrix b1 b 2 b 3 b 4 b 5 b 6 V R n1 S 1 1 1 0 0 0 W n2S 0 - 1 0 - 1 1 0 W n 3SS- 1 0 0 0 - 1 - 1WW n 4S 0 0 - 1 1 0 1 W X T In option (C)

nodia E = AV V R S1 1 1 0 0 0W S 0 -1 0 -1 1 0 W T T 8e1 e2 e 3 e 4B = S- 1 0 0 0 - 1 - 1W8V1 V2 -- V6B W S S 0 0 -1 1 0 1 W X R V RT V Se1W S V1 + V2 + V3 W Se2W S- V2 - V4 + V5W Se W = S- V - V - V W which is true. 5 6W S 3W S 1 Se 4W S- V3 + V4 + V6W T X T X

SOL 1.78

Correct option is (A). Assume a Gaussian surface inside the sphere (x < R)

From gauss law y = Q enclosed = Q enclosed =

# D : ds = Q enclosed 3 Q 4 pr3 = Qr # 3 4 3 R3 3 pR


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So, or SOL 1.79

Qr3 R3 Qr3 Q r D # 4pr2 = 3 = 4pe0 R3 R

# D : ds

Page 67

=

a D = e0 E

Correct option is (D). Inductance is given as L =

m0 N2 A l

4p # 10- 7 # (400) 2 # (16 # 10- 4) (1 # 10- 3) = 321.6 mH ` XL = 2pfL V = IXL = 230 2pfL 230 = = 2.28 A 2 # 3.14 # 50 # 321.6 # 10- 3 =

SOL 1.80

SOL 1.81

nodia Correct option is (A). Energy stored is inductor E = 1 LI2 2 = 1 # 321.6 # 10- 3 # (2.28) 2 2 Force required to reduce the air gap of length 1 mm is F = E = 0.835- 3 = 835 N l 1 # 10 Correct option is (D). Thevenin voltage:

Vth = I (R + ZL + ZC ) = 1+0c [1 + 2j - j] = 1 (1 + j) =

2 +45% V

Thevenin impedance:

Zth = R + ZL + ZC = 1 + 2j - j = (1 + j) W


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Chapter 1

Correct option is (A). In the given circuit

Output voltage vo = Avi = 106 # 1 mV = 1 V Input impedance Zi = vi = vi = 3 0 ii Output impedance Zo = vo = Avi = Ro = 10 W io io

SOL 1.83

nodia

Correct option is (D). All sources present in the circuit are DC sources, so all inductors behaves as short circuit and all capacitors as open circuit Equivalent circuit is

Voltage across R 3 is 5 = I1 R 3 5 = I1 (1) I1 = 5 A By applying KCL, current through voltage source

(current through R 3 )

1 + I 2 = I1 I2 = 5 - 1 = 4 A SOL 1.84

Correct option is (). Given Two port network can be described in terms of h-parametrs only.

SOL 1.85

Correct option is (A). At resonance reactance of the circuit would be zero and voltage across inductor and capacitor would be equal VL = VC At resonance impedance of the circuit Z R = R1 + R 2


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Current

IR = V1 +0c R1 + R 2

Voltage

V2 = IR R2 + j (VL - VC ) V2 = V1 +0c R2 R1 + R 2

Page 69

Voltage across capacitor 1 I jw C # R = 1 # VR +0c R1 + R 2 jwC = VR + - 90c wC (R1 + R2)

VC =

So phasor diagram is

SOL 1.86

nodia Correct option is (B). This is a second order LC circuit shown below

Capacitor current is given as iC (t) = C

dvc (t) dt

Taking Laplace transform IC (s) = CsV (s) - V (0), V (0) " initial voltage Current in inductor iL (t) = 1 L

# vc (t) dt

V (s) IL (s) = 1 L s For t > 0 , applying KCL(in s-domain) IC (s) + IL (s) = 0 V (s) =0 CsV (s) - V (0) + 1 L s 1 2 :s + LCs D V (s) = Vo V (s) = Vo 2 s 2 , s + w0 Taking inverse Laplace transformation v (t) = Vo cos wo t ,

a w20 = 1 LC t>0


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Chapter 1

Correct option is (B). Power dissipated in heater when AC source is connected 2 P = 2.3 kW = V rms R (230) 2 R

2.3 # 103 =

R = 23 W (Resistance of heater) Now it is connected with a square wave source of 400 V peak to peak Power dissipated is 2

P = V rms , R =

SOL 1.88

(200) 2 = 1.739 kW 23

nodia

(Divergence of electric field intensity is non-Zero)

4: B = 0

(Divergence of magnetic field intensity is zero)

Correct option is (C). Current in the circuit

I =

Or SOL 1.90

100 =8 A R + (10 || 10)

(given)

100 = 8 R+5 R = 60 = 7.5 W 8

Correct option is (A). Rms value is given as mrms = =

SOL 1.91

Vrms = Vp =200 (for square wave)

Correct option is (D). From maxwell’s first equation 4: D = rv r 4: E = v e Maxwell’s fourth equation

SOL 1.89

Vp - p = 400 V & Vp = 200 V

32 +

(4) 2 2

9 + 8 = 17 V

Correct option is (D). Writing KVL in input and output loops V1 - (i1 + i2) Z1 = 0 V1 = Z1 i1 + Z1 i2 Similarly V2 - i2 Z2 - (i1 + i2) Z1 = 0 V2 = Z1 i1 + (Z1 + Z2) i2 From equation (1) and (2) Z -matrix is given as Z1 Z1 Z => Z1 Z1 + Z2H

...(1)

...(2)


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Page 71

Correct option is (B). In final steady state the capacitor will be completely charged and behaves as an open circuit

Steady state voltage across capacitor vc (3) = 20 (10) 10 + 10

nodia = 10 V

SOL 1.93

Correct option is (D). We know that divergence of the curl of any vector field is zero 4 (4 # E) = 0

SOL 1.94

Correct option is (A). When the switch is at position 1, current in inductor is given as

120 = 2 A 20 + 40 At t = 0 , when switch is moved to position 1,inductor current does not change simultaneously so iL (0-) =

iL (0+) = iL (0-)=2 A Voltage across inductor at t = 0+ vL (0+) = 120 V By applying KVL in loop 120 = 2 (40 + R + 20) 120 = 120 + R R = 0W


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Chapter 1

Correct option is (C). Let stored energy and dissipated energy are E1 and E2 respectively. Then Current i 22 = E2 = 0.95 E1 i 12 i2 = 0.95 i1 = 0.97i1 Current at any time t, when the switch is in position (2) is given by R

60

i (t) = i1 e- L t = 2e- 10 t = 2e- 6t After 95% of energy dissipated current remaining in the circuit is i = 2 - 2 # 0.97 = 0.05 A 0.05 = 2e- 6t t . 0.50 sec

So,

SOL 1.96

nodia Correct option is (C). At f1 = 100 Hz, voltage drop across R and L is mRMS mRMS = Vin .R R + jw1 L V (jw L) = in 1 R + jw1 L

So, R = w1 L At f2 = 50 Hz, voltage drop across R mlRMS = Vin .R R + jw2 L

mRMS R + jw2 L = = R + jw1 L mlRMS = = = mlRMS = SOL 1.97

w12 L2 + w22 L2 , w12 L2 + w12 L2

R2 + w22 L2 R2 + w12 L2 R = w1 L

f 12 + f 22 w12 + w22 = 2f 12 2w12 (100) 2 + (50) 2 = 5 8 2 (100) 2 8m 5 RMS

Correct option is (A). In the circuit I B = IR +0c + Iy +120c I B2 = I R2 + I y2 + 2IR Iy cos b 120c l 2 Since so,

= I R2 + I y2 + IR Iy I R = Iy I B2 = I R2 + I R2 + I R2 = 3I R2 IB = 3 IR =

3 Iy

IR: Iy: IB = 1: 1: 3


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Page 73

Correct option is (C). Switch was opened before t = 0 , so current in inductor for t < 0

iL (0-) = 10 = 1 A 10 Inductor current does not change simultaneously so at t = 0 when switch is closed current remains same

nodia iL (0+) = iL (0-)=1 A

SOL 1.99

SOL 1.100

Correct option is (A). Electric field inside a conductor (metal) is zero. In dielectric charge distribution os constant so electric field remains constant from x1 to x2 . In semiconductor electric field varies linearly with charge density. Correct option is (A). Thevenin voltage:

Nodal analysis at P Vth - 4 + Vth = 0 10 10 2Vth - 4 = 0 Vth = 2 V Thevenin resistance:

Rth = 10 W || 10 W = 5 W SOL 1.101

Correct option is (D). Resonance will occur only when Z is capacitive, in parallel resonance condition, suseptance of circuit should be zero.


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Chapter 1

1 + j wC = 0 j wL 1 - w2 LC = 0 1 (resonant frequency) LC 1 1 C = 2 = = 0.05 m F 2 4 # p # (500) 2 # 2 wL w=

SOL 1.102

Correct option is (D). Here two capacitor C1 and C2 are connected in series so equivalent Capacitance is Ceq = C1 C2 C1 + C 2 8.85 # 10- 12 # 4 (400 # 10- 3) 2 C1 = e0 er1 A = d1 6 # 10- 3

nodia Similarly

- 12 16 # 10- 2 = 94.4 10- 11 F = 8.85 # 10 # 4 # # -3 6 # 10

8.85 # 10- 12 # 2 # (400 # 10- 3) 2 C2 = e0 er2 A = d2 8 # 10- 3 - 12 16 # 10- 12 = 8.85 # 10 # 2 # -3 8 # 10 - 11 = 35.4 # 10 F

- 11 - 11 = 25.74 # 10- 11 - 257 pF Ceq = 94.4 # 10 # 35.4 #-10 11 (94.4 + 35.4) # 10

SOL 1.103

Correct option is (C). Inductance of the Solenoid is given as m N2 A L = 0 l Where

SOL 1.104

A " are of Solenoid l " length 4p # 10- 7 # (3000) 2 # p (30 # 10- 3) 2 L = (1000 # 10- 3) = 31.94 # 10- 3 H - 32 mH

Correct option is (C). In the circuit

Voltage So,

VA = (2 + 1) # 6 = 18 Volt 2 = E - VA 6


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Page 75

2 = E - 18 6 E = 12 + 18 = 30 V SOL 1.105

SOL 1.106

Correct option is (A). Delta to star (T - Y) conversions is Rb Rc R1 = Ra + Rb + Rc Ra Rc R2 = Ra + Rb + Rc Ra Rb R3 = Ra + Rb + Rc

given as = 10 # 10 = 2.5 W 20 + 10 + 10 = 20 # 10 = 5 W 20 + 10 + 10 = 20 # 10 = 5 W 20 + 10 + 10

nodia Correct option is (D). For parallel circuit

I = E = EYeq Zeq

Yeq " Equivalent admittance of the circuit Yeq = YR + YL + YC = (0.5 + j0) + (0 - j1.5) + (0 + j0.3)

= 0.5 - j1.2 So, current I = 10 (0.5 - j1.2) = (5 - j12) A

SOL 1.107


100 (10 || R) 10 + (10 || R) # 100 10R = f 10 + 10R pb 10 + R l 10 + R = 1000R = 50R 100 + 20R 5+R Current in R W resistor 2 = VA R 2 = 50R R (5 + R) Voltage

or SOL 1.108

VA =

R = 20 W

Correct option is (A). Since capacitor initially has a charge of 10 coulomb, therefore


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Chapter 1

vc (0) " initial voltage across capacitor Q 0 = Cvc (0) 10 = 0.5vc (0) vc (0) = 10 = 20 V 0.5 When switch S is closed, in steady state capacitor will be charged completely and capacitor voltage is vc (3) = 100 V At any time t transient response is t

vc (t) = vc (3) + [vc (0) - vc (3)] e- RC t

vc (t) = 100 + (20 - 100) e- 2 # 0.5 = 100 - 80e- t

nodia Current in the circuit

i (t) = C dvc = C d [100 - 80e- t] dt dt = C # 80e- t

= 0.5 # 80e- t

= 40e- t

At t = 1 sec,

i (t) = 40e- 1 = 14.71 A

SOL 1.109

Correct option is (D). Total current in the wire

I = 10 + 20 sin wt 102 +

Irms =

(20) 2 2

= 100 + 200 =

300 = 17.32 A

SOL 1.110

Correct option is (D). From Z to Y parameter conversion Z11 Z12 - 1 Y11 Y12 >Y Y H = >Z Z H 21 22 21 22 0.6 - 0.2 Y11 Y12 So, >Y Y H = 0.150 >- 0.2 0.9 H 12 22 Y22 = 0.9 = 1.8 0.50

SOL 1.111

Correct option is (C). Energy absorbed by the inductor coil is given as t

EL =

# Pdt 0

Where power

P = VI = I bL dI l dt t

So,

EL =

dt # LIb dI dt l

0


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Page 77

For0 # t # 4 sec 4

dt # Ib dI dt l

EL = 2

0

a dI = 3, 0 # t # 2 , # # * dt 0 2 = 0, 2 < t < 4 2 = 6 # I.dt =6(area under the curve i (t) - t ) 2

=2

4

I (3) dt + 2

I (0) dt

0

= 6 # 1 # 2 # 6 = 36 J 2 Energy absorbed by 1 W resistor is t

ER =

#

2

I2 Rdt =

# (3t) 2 # 1dt + 0

0

4

# (6) 2 dt 2

I = 3t, 0 # t # 2 ) = 6A 2 # t # 4

nodia 3 2

= 9 # :t D + 36[t]2 = 24 + 72 =96 J 3 0 Total energy absorbed in 4 sec E = EL + ER = 36 + 96 = 132 J

SOL 1.112

4

Correct option is (B). Applying KCL at center node

iL = iC + 1 + 2 iL = iC + 3 iC =- C dvc =- 1 d [4 sin 2t] dt dt so Voltage across inductor

=- 8 cos 2t iL =- 8 cos 2t + 3

(current through inductor)

vL = L diL = 2 # d [3 - 8 cos 2t] = 32 sin 2t dt dt SOL 1.113

Correct option is (A). Thevenin impedance can be obtain as following


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Chapter 1

Zth = Z 3 + (Z1 || Z2) Z1 = 10+ - 60c

Given that

= 10 c

1-

2

3j

m = 5 (1 -

3 j)

3j

m = 5 (1 +

3 j)

Z2 = 10+60c = 10 c

1+

2

Z 3 = 50+53.13c 3 + 4j = 50 b = 10 (3 + 4j) 5 l 5 (1 - 3j) 5 (1 + 3 j) 5 (1 - 3 j) + 5 (1 + 3 j) 25 (1 + 3) = 10 (3 + 4j) + 10

Zth = 10 (3 + 4j) +

So,

nodia = 30 + 40j + 10 = 40 + 40j

Zth = 40 2 +45c W

SOL 1.114

Correct option is (A). Due to the first conductor carrying + I current, magnetic field intensity at point P is H 1 = I Y (Direction is determined using right hand rule) 2pd Similarly due to second conductor carrying - I current, magnetic field intensity is H 2 = - I (- Y) = I Y 2pd 2pd Total magnetic field intensity at point P. H = H1 + H 2 = I Y+ I Y = I Y 2pd 2pd pd

SOL 1.115

Correct option is ( ).

SOL 1.116

Correct option is (C). Given that magnitudes of VL and VC are twice of VR VL = VC = 2VR (Circuit is at resonance) Voltage across inductor VL = iR # jwL Current iR at resonance % iR = 5+0 = 5 = 1 A 5 R so,

VL = wL = 2VR wL = 2 # 5 2 # p # 50 # L = 10 L = 10 = 31.8 mH 314

VR = 5 V, at resonance


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Page 79

Correct option is (C). Applying nodal analysis in the circuit At node P 2 + VP - 10 + VP = 0 2 8 16 + 4VP - 40 + VP = 0 5VP - 24 = 0 VP = 24 Volt 5 At node Q V - 10 VQ - 0 + 2= Q 6 4 24 = 3VQ - 30 + 2VQ 5VQ - 54 = 0 VQ = 54 V 5 Potential difference between P-Q

nodia VPQ = VP - VQ = 24 - 54 =- 6 V 5 5

SOL 1.118

Correct option is (D). First obtain equivalent Thevenin circuit across load RL

Thevenin voltage Vth - 110+0c + Vth - 90+0c 0 = 6 + 8j 6 + 8j 2Vth - 200+0c = 0 Vth = 100+0c V Thevenin impedance

Zth = (6 + 8j) W || (6 + 8j) W = (3 + 4j) W For maximum power transfer RL = Zth =

32 + 42 = 5 W


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Chapter 1

Power in load 2 P = ieff RL 2 100 P = 5 3 + 4j + 5 #

=

(100) 2 5 80 #

= 625 Watt

SOL 1.119

nodia Correct option is (D). Applying mesh analysis in the circuit

I1 = 10 A, I2 =- 5 A Current in 2 W resistor I2W = I1 - (- I2) = 10 - (- 5) = 15 A So, voltage VA = 15 # 2 = 30 Volt Now we can easily find out current in all branches as following

Current in resistor R is 5 A 5 = 100 - 40 R R = 60 = 12 W 5


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Page 81

Correct option is (B). Before t = 0 , the switch was opened so current in inductor and voltage across capacitor for t < 0 is zero vc (0-) = 0 , iL (0- ) = 0 at t = 0 , when the switch is closed, inductor current and capacitor voltage does not change simultaneously so vc (0+) = vc (0-) = 0 , iL (0+) = iL (0-) = 0 At t = 0+ the circuit is

nodia Simplified circuit

Voltage across inductor (at t = 0+ ) vL (0+) = 10 # 2 3+2 = 4 Volt SOL 1.121

Correct option is (D). Given that E1 = h11 I1 + h12 E2 and I2 = h21 I1 + h22 E2 Parameter h12 is given as h12 = E1 E2 I = 0 (open circuit) 1

At node A E A - E1 + E A - E 2 + E A = 0 2 2 4 5EA = 2E1 + 2E2

...(1)


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Chapter 1

Similarly E1 - E A + E1 = 0 2 2 2E1 = EA

...(2)

From (1) and (2) 5 (2E1) = 2E1 + 2E2 4 8E1 = 2E2 h12 = E1 = 1 4 E2 SOL 1.122

Correct option is (B). VPQ = VP - VQ KQ KQ = OP OQ

nodia

9 10- 9 - 9 # 109 # 1 # 10- 9 = 9 # 10 # 1 # -3 40 # 10 20 # 10- 3 = 9 # 103 : 1 - 1 D =- 225 Volt 40 20

SOL 1.123

Correct option is (D). Energy stored in Capacitor is E = 1 CV2 2

- 12 -6 # 10 C = e0 A = 8.85 # 10 # 100 3 d 0.1 # 10

= 8.85 # 10- 12 F E = 1 # 8.85 # 10- 12 # (100) 2 2 = 44.3 nJ SOL 1.124

Correct option is (B). The figure is as shown below

The Capacitor shown in Figure is made up of two capacitor C1 and C2 connected in series. C1 = e0 er1 A , C2 = e0 er2 A t1 t2 Since C1 and C2 are in series charge on both capacitor is same. Q1 = Q 2 C1 (100 - V) = C2 V (Let V is the voltage of foil)


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Page 83

e0 er1 A (100 - V) = e0 er2 A V t2 t1 3 (100 - V) = 4 V 0.5 1 300 - 3V = 2V 300 = 5V & V = 60 Volt

************

nodia


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Signals and Systems

Chapter 2

CHAPTER 2 SIGNALS AND SYSTEMS

2014 EE01 EE SP 2.1

EE SP 2.2

ONE MARK

Let X ^s h = 2 3s + 5 be the Laplace Transform of a signal x ^ t h. Then, x ^0+h s + 10s + 21 is (A) 0

(B) 3

(C) 5

(D) 21

nodia

For a periodic square wave, which one of the following statements is TRUE ? (A) The Fourier series coefficients do not exist. (B) The Fourier series coefficients exist but the reconstruction converges at no point. (C) The Fourier series coefficients exist and the reconstruction converges at most points. (D) The Fourier series coefficients exist and the reconstruction converges at every point. 2014 EE02

EE SP 2.3

TWO MARKS

Let f ^ t h be a continuous time signal and let F ^wh be its Fourier Transform defined by F ^w h = Define g ^ t h by

# g^t h = #

3

-3 3

-3

f ^ t h e-jwt dt F û h e-jut du

What is the relationship between f ^ t h and g ^ t h ? (A) g ^ t h would always be proportional to f ^ t h

(B) g ^ t h would be proportional to f ^ t h if f ^ t h is an even function (C) g ^ t h would be proportional to f ^ t h only if f ^ t h is a sinusoidal function (D) g ^ t h would never be proportional to f ^ t h EE SP 2.4

The function shown in the figure can be represented as


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Signals and Systems

Page 85

^t - T h ^t - 2T h u ^t - 2T h u ^t - T h T T (B) u ^ t h + t u ^t - T h - t u ^t - 2T h T T ^t - T h ^t - 2T h (C) u ^ t h - u ^t - T h + u^t h u^t h T T ^t - T h ^t - 2T h (D) u ^ t h + u ^t - 2T h u ^t - T h - 2 T T (A) u ^ t h - u ^t - T h +

2014 EE02 EE SP 2.5

EE SP 2.6

ONE MARK

Consider an LTI system with transfer function 1 H ^s h = s ^s + 4h If the input to the system is cos ^3t h and the steady state output is A sin ^3t + ah , then the value of A is (A) 1/30 (B) 1/15 (C) 3/4 (D) 4/3

nodia

Consider an LTI system with impulse response h ^ t h = e-5t u ^ t h. If the output of the system is y ^ t h = e-3t u ^ t h - e-5t u ^ t h then the input, x ^ t h, is given by (A) e-3t u ^ t h (B) 2e-3t u ^ t h (C) e-5t u ^ t h (D) 2e-5t u ^ t h 2014 EE02

EE SP 2.7

EE SP 2.8

TWO MARKS

A discrete system is represented by the difference equation a a - 1 X1 ^k h X1 ^k + 1h > H = >a + 1 a H> H X2 ^k + 1h X2 ^k h It has initial conditions X1 ^0 h = 1; X2 ^0 h = 0 . The pole locations of the system for a = 1, are (A) 1 ! j0 (B) - 1 ! j0 (C) ! 1 + j0 (D) 0 ! j1 An input signal x ^ t h = 2 + 5 sin ^100pt h is sampled with a sampling frequency of 400 Hz and applied to the system whose transfer function is represented by -N Y ^z h = 1 c 1 - z -1 m N 1-z X ^z h where, N represents the number of samples per cycle. The output y ^n h of the system under steady state is (A) 0 (B) 1 (C) 2 (D) 5


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Signals and Systems

Chapter 2

2014 EE03 EE SP 2.9

ONE MARK

A function f ^ t h is shown in the figure.

The Fourier transform F ^wh of f ^ t h is (A) real and even function of w (B) real and odd function of w (C) imaginary and odd function of w

nodia (D) imaginary and even function of w

EE SP 2.10

A signal is represented by

1 t <1 x^t h = * 0 t >1 The Fourier transform of the convolved signal y ^ t h = x ^2t h * x ^t/2h is

(A) 42 sin a w k sin ^2w h 2 w (C) 42 sin ^2wh w

EE SP 2.11

(B) 42 sin a w k 2 w (D) 42 sin2 w w

For the signal f ^ t h = 3 sin 8pt + 6 sin 12pt + sin 14pt , the minimum sampling frequency (in Hz) satisfying the Nyquist criterion is _____. 2014 EE03

EE SP 2.12

TWO MARKS

A continuous-time LTI system with system function H ^wh has the following polezero plot. For this system, which of the alternatives is TRUE ?

(A) H ^0 h (B) H ^wh (C) H ^0 h (D) H ^wh

> ^w h ; w > 0 has multiple maxima, at w 1 and w 2 < H ^w h ; w > 0 = constant; - 3 < w < 3


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Signals and Systems

A sinusoid x ^ t h of unknown frequency is sampled by an impulse train of period 20 ms. The resulting sample train is next applied to an ideal lowpass filter with cutoff at 25 Hz . The filter output is seen to be a sinusoid of frequency 20 Hz. This means that x ^ t h (A) 10 Hz (B) 60 Hz (C) 30 Hz

EE SP 2.14

Page 87

(D) 90 Hz

A differentiable non constant even function x ^ t h has a derivative y ^ t h, and their respective Fourier Transforms are X ^wh and Y ^wh. Which of the following statments is TRUE ? (A) X ^wh and Y ^wh are both real (B) X ^wh is real and Y ^wh is imaginary (C) X ^wh and Y ^wh are both imaginary (D) X ^wh is imaginary and Y ^wh is real

nodia YEAR 2013

EE SP 2.15

ONE MARK

A band-limited signal with a maximum frequency of 5 kHz is to be sampled. According to the sampling theorem, the sampling frequency which is not valid is (A) 5 kHZ (B) 12 kHz (C) 15 kHz

(D) 20 kHz EE SP 2.16

EE SP 2.17

EE SP 2.18

For a periodic signal v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p/4h, the fundamental frequency in rad/s (A) 100 (B) 300 (C) 500 (D) 1500 Two systems with impulse responses h1 ^ t h and h2 ^ t h are connected in cascade. Then the overall impulse response of the cascaded system is given by (A) product of h1 ^ t h and h2 ^ t h (B) sum of h1 ^ t h and h2 ^ t h (C) convolution of h1 ^ t h and h2 ^ t h (D) subtraction of h2 ^ t h from h1 ^ t h Which one of the following statements is NOT TRUE for a continuous time causal and stable LTI system? (A) All the poles of the system must lie on the left side of the jw axis (B) Zeros of the system can lie anywhere in the s-plane (C) All the poles must lie within s = 1 (D) All the roots of the characteristic equation must be located on the left side of the jw axis.

EE SP 2.19

The impulse response of a system is h ^ t h = tu ^ t h. For an input u ^t - 1h, the


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Signals and Systems

output is 2 (A) t u ^ t h 2 (C)

Chapter 2

(B)

^t - 1h2 u ^t - 1h 2

t ^t - 1h u ^t - 1h 2

2 (D) t - 1 u ^t - 1h 2


TWO MARKS

The impulse response of a continuous time system is given by h ^ t h = d ^t - 1h + d ^t - 3h . The value of the step response at t = 2 is (A) 0 (B) 1 (C) 2

(D) 3

nodia YEAR 2012

EE SP 2.21

EE SP 2.22

ONE MARK

If x [n] = (1/3) n - (1/2) n u [n], then the region of convergence (ROC) of its z -transform in the z -plane will be (A) 1 < z < 3 (B) 1 < z < 1 3 3 2 (C) 1 < z < 3 (D) 1 < z 2 3

The unilateral Laplace transform of f (t) is 2 1 . The unilateral Laplace s +s+1 transform of tf (t) is (A) - 2 s (B) - 2 2s + 1 2 (s + s + 1) 2 (s + s + 1) (C) 2 s (D) 2 2s + 1 2 (s + s + 1) 2 (s + s + 1) YEAR 2012

TWO MARKS

EE SP 2.23

Let y [n] denote the convolution of h [n] and g [n], where h [n] = (1/2) n u [n] and g [n] is a causal sequence. If y [0] = 1 and y [1] = 1/2, then g [1] equals (A) 0 (B) 1/2 (C) 1 (D) 3/2

EE SP 2.24

The Fourier transform of a signal h (t) is H (jw) = (2 cos w) (sin 2w) /w . The value of h (0) is (A) 1/4 (B) 1/2 (C) 1 (D) 2

EE SP 2.25

The input x (t) and output y (t) of a system are related as y (t) = . The system is

t

# x (t) cos (3t) dt -3


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Signals and Systems

Page 89

(A) time-invariant and stable (B) stable and not time-invariant (C) time-invariant and not stable (D) not time-invariant and not stable


ONE MARK

The Fourier series expansion f (t) = a 0 +

3

/ an cos nwt + bn sin nwt

of

n=1

the periodic signal shown below will contain the following nonzero terms

nodia (A) a 0 and bn, n = 1, 3, 5, ...3 (B) a 0 and an, n = 1, 2, 3, ...3 (C) a 0 an and bn, n = 1, 2, 3, ...3 (D) a 0 and an n = 1, 3, 5, ...3

EE SP 2.27

Given two continuous time signals x (t) = e-t and y (t) = e-2t which exist for t > 0 , the convolution z (t) = x (t) * y (t) is (A) e-t - e-2t (B) e-3t (C) e+t (D) e-t + e-2t

YEAR 2011

TWO MARKS

EE SP 2.28

Let the Laplace transform of a function f (t) which exists for t > 0 be F1 (s) and the Laplace transform of its delayed version f (t - t) be F2 (s). Let F1 * (s) be the complex conjugate of F1 (s) with the Laplace variable set s = s + jw . If F (s) F1 * (s) , then the inverse Laplace transform of G (s) is an ideal G (s) = 2 F1 (s) 2 (A) impulse d (t) (B) delayed impulse d (t - t) (C) step function u (t) (D) delayed step function u (t - t)

EE SP 2.29

The response h (t) of a linear time invariant system to an impulse d (t), under initially relaxed condition is h (t) = e-t + e-2t . The response of this system for a


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Signals and Systems

Chapter 2

unit step input u (t) is (A) u (t) + e-t + e-2t (B) (e-t + e-2t) u (t) (C) (1.5 - e-t - 0.5e-2t) u (t) (D) e-t d (t) + e-2t u (t) YEAR 2010 EE SP 2.30

ONE MARK

For the system 2/ (s + 1), the approximate time taken for a step response to reach 98% of the final value is (A) 1 s (B) 2 s

nodia (C) 4 s

(D) 8 s

EE SP 2.31

The period of the signal x (t) = 8 sin `0.8pt + p j is 4 (A) 0.4p s (B) 0.8p s (C) 1.25 s (D) 2.5 s

EE SP 2.32

The system represented by the input-output relationship y (t) = (A) Linear and causal

5t

# x (t) dt, t > 0 is -3

(B) Linear but not causal (C) Causal but not linear (D) Neither liner nor causal EE SP 2.33

The second harmonic component of the periodic waveform given in the figure has an amplitude of

(A) 0 (B) 1 (C) 2/p (D)

5


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Signals and Systems

Page 91


TWO MARKS

x (t) is a positive rectangular pulse from t =- 1 to t =+ 1 with unit height as shown in the figure. The value of transform of x (t)} is.

(A) 2

3

X (w) 2 dw " where X (w) is the Fourier

(B) 2p (D) 4p

nodia (C) 4

EE SP 2.35

#- 3

Given the finite length input x [n] and the corresponding finite length output y [n] of an LTI system as shown below, the impulse response h [n] of the system is

(A) h [n] = {1, 0, 0, 1} (C) h [n] = {1, 1, 1, 1} -

(B) h [n] = {1, 0, 1} (D) h [n] = {1, 1, 1} -

Common Data Questions Q. 36 and 37 Given f (t) and g (t) as show below

EE SP 2.36

EE SP 2.37

g (t) can be expressed as (A) g (t) = f (2t - 3) (C) g (t) = f`2t - 3 j 2

(B) g (t) = f` t - 3j 2 (D) g (t) = f` t - 3 j 2 2

The Laplace transform of g (t) is (A) 1 (e3s - e5s) s

(B) 1 (e - 5s - e - 3s) s

- 3s (C) e (1 - e - 2s) s

(D) 1 (e5s - e3s) s


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Signals and Systems

Chapter 2


ONE MARK

A Linear Time Invariant system with an impulse response h (t) produces output y (t) when input x (t) is applied. When the input x (t - t) is applied to a system with impulse response h (t - t), the output will be (A) y (t) (B) y (2 (t - t)) (C) y (t - t) (D) y (t - 2t) YEAR 2009

EE SP 2.39

TWO MARKS

nodia

A cascade of three Linear Time Invariant systems is causal and unstable. From this, we conclude that (A) each system in the cascade is individually causal and unstable (B) at least on system is unstable and at least one system is causal (C) at least one system is causal and all systems are unstable (D) the majority are unstable and the majority are causal

EE SP 2.40

The Fourier Series coefficients of a periodic signal x (t) expressed as x (t) = k3=- 3 ak e j2pkt/T are given by a- 2 = 2 - j1, a- 1 = 0.5 + j0.2 , a 0 = j2 ,

/

a 1 = 0.5 - j0.2 , a 2 = 2 + j1 and ak = 0 for k > 2 Which of the following is true ? (A) x (t) has finite energy because only finitely many coefficients are non-zero (B) x (t) has zero average value because it is periodic (C) The imaginary part of x (t) is constant (D) The real part of x (t) is even

EE SP 2.41

The z-transform of a signal x [n] is given by 4z - 3 + 3z - 1 + 2 - 6z2 + 2z3 It is applied to a system, with a transfer function H (z) = 3z - 1 - 2 Let the output be y [n]. Which of the following is true ? (A) y [n] is non causal with finite support (B) y [n] is causal with infinite support (C) y [n] = 0; n > 3 (D) Re [Y (z)] z = e =- Re [Y (z)] z = e ji

- ji

Im [Y (z)] z = e = Im [Y (z)] z = e ; - p # q < p ji

- ji


ONE MARK

The impulse response of a causal linear time-invariant system is given as h (t). Now consider the following two statements : Statement (I): Principle of superposition holds Statement (II): h (t) = 0 for t < 0 Which one of the following statements is correct ? (A) Statements (I) is correct and statement (II) is wrong (B) Statements (II) is correct and statement (I) is wrong


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Signals and Systems

Page 93

(C) Both Statement (I) and Statement (II) are wrong (D) Both Statement (I) and Statement (II) are correct EE SP 2.43

A signal e - at sin (wt) is the input to a real Linear Time Invariant system. Given K and f are constants, the output of the system will be of the form Ke - bt sin (vt + f) where (A) b need not be equal to a but v equal to w (B) v need not be equal to w but b equal to a (C) b equal to a and v equal to w (D) b need not be equal to a and v need not be equal to w

nodia YEAR 2008

EE SP 2.44

TWO MARKS

z Given X (z) = with z > a , the residue of X (z) zn - 1 at z = a for n $ 0 2 (z - a) will be (A) an - 1 (B) an (D) nan - 1

(C) nan

EE SP 2.45

A system with x (t) and output y (t) is defined by the input-output relation : y (t) =

- 2t

#- 3x (t) dt

The system will be (A) Casual, time-invariant and unstable (B) Casual, time-invariant and stable (C) non-casual, time-invariant and unstable (D) non-casual, time-variant and unstable EE SP 2.46

A signal x (t) = sinc (at) where a is a real constant ^sinc (x) = px h is the input to a Linear Time Invariant system whose impulse response h (t) = sinc (bt), where b is a real constant. If min (a, b) denotes the minimum of a and b and similarly, max (a, b) denotes the maximum of a and b, and K is a constant, which one of the following statements is true about the output of the system ? (A) It will be of the form Ksinc (gt) where g = min (a, b) (B) It will be of the form Ksinc (gt) where g = max (a, b) (C) It will be of the form Ksinc (at) (D) It can not be a sinc type of signal sin (px)

EE SP 2.47

Let x (t) be a periodic signal with time period T , Let y (t) = x (t - t0) + x (t + t0) for some t0 . The Fourier Series coefficients of y (t) are denoted by bk . If bk = 0 for all odd k , then t0 can be equal to (A) T/8 (B) T/4 (C) T/2 (D) 2T

EE SP 2.48

H (z) is a transfer function of a real system. When a signal x [n] = (1 + j) n is the


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Signals and Systems

Chapter 2

input to such a system, the output is zero. Further, the Region of convergence (ROC) of ^1 - 12 z - 1h H(z) is the entire Z-plane (except z = 0 ). It can then be inferred that H (z) can have a minimum of (A) one pole and one zero (B) one pole and two zeros (C) two poles and one zero D) two poles and two zeros EE SP 2.49

Let x (t) = rect^t - 12 h (where rect (x) = 1 for - 12 # x #

1 2

and zero otherwise.

sin (px)

If sinc (x) = px , then the FTof x (t) + x (- t) will be given by (B) 2 sinc` w j (A) sinc` w j 2p 2p (C) 2 sinc` w j cos ` w j (D) sinc` w j sin ` w j 2p 2 2p 2

EE SP 2.50

nodia

Given a sequence x [n], to generate the sequence y [n] = x [3 - 4n], which one of the following procedures would be correct ? (A) First delay x (n) by 3 samples to generate z1 [n], then pick every 4th sample of z1 [n] to generate z2 [n], and than finally time reverse z2 [n] to obtain y [n]. (B) First advance x [n] by 3 samples to generate z1 [n], then pick every 4th sample of z1 [n] to generate z2 [n], and then finally time reverse z2 [n] to obtain y [n] (C) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n] to obtain v2 [n], and finally advance v2 [n] by 3 samples to obtain y [n] (D) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n] to obtain v2 [n], and finally delay v2 [n] by 3 samples to obtain y [n] YEAR 2007

EE SP 2.51

ONE MARK

Let a signal a1 sin (w1 t + f) be applied to a stable linear time variant system. Let the corresponding steady state output be represented as a2 F (w2 t + f2). Then which of the following statement is true? (A) F is not necessarily a “Sine” or “Cosine” function but must be periodic with w1 = w2 . (B) F must be a “Sine” or “Cosine” function with a1 = a2 (C) F must be a “Sine” function with w1 = w2 and f1 = f2 (D) F must be a “Sine” or “Cosine” function with w1 = w2

EE SP 2.52

The frequency spectrum of a signal is shown in the figure. If this is ideally sampled at intervals of 1 ms, then the frequency spectrum of the sampled signal will be


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Signals and Systems

Page 95

nodia YEAR 2007 EE SP 2.53

TWO MARKS

A signal x (t) is given by 1, - T/4 < t # 3T/4 x (t) = *- 1, 3T/4 < t # 7T/4 - x (t + T) Which among the following gives the fundamental fourier term of x (t) ? (B) p cos ` pt + p j (A) 4 cos ` pt - p j 4 4 2T p T 4 (C) 4 sin ` pt - p j (D) p sin ` pt + p j 4 p T 4 4 2T

EE SP 2.54

Consider the discrete-time system shown in the figure where the impulse response of G (z) is g (0) = 0, g (1) = g (2) = 1, g (3) = g (4) = g = 0


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Signals and Systems

Chapter 2

This system is stable for range of values of K (A) [- 1, 12 ] (B) [- 1, 1] (D) [- 12 , 2] (C) [- 12 , 1]

Common Data Questions Q. 55 and 56 A signal is processed by a causal filter with transfer function G (s) EE SP 2.55

nodia For a distortion free output signal wave form, G (s) must (A) provides zero phase shift for all frequency (B) provides constant phase shift for all frequency

(C) provides linear phase shift that is proportional to frequency

(D) provides a phase shift that is inversely proportional to frequency

EE SP 2.56

G (z) = az - 1 + bz - 3 is a low pass digital filter with a phase characteristics same as that of the above question if (A) a = b (B) a =- b (D) a = b(- 1/3)

(C) a = b(1/3) EE SP 2.57

If u (t), r (t) denote the unit step and unit ramp functions respectively and u (t) * r (t) their convolution, then the function u (t + 1) * r (t - 2) is given by (A) 12 (t - 1) u (t - 1) (B) 12 (t - 1) u (t - 2) (C)

EE SP 2.58

1 2

(t - 1) 2 u (t - 1)

(D) None of the above

X (z) = 1 - 3z - 1, Y (z) = 1 + 2z - 2 are Z transforms of two signals x [n], y [n] respectively. A linear time invariant system has the impulse response h [n] defined by these two signals as h [n] = x [n - 1] * y [n] where * denotes discrete time convolution. Then the output of the system for the input d [n - 1] (A) has Z-transform z - 1 X (z) Y (z) (B) equals d [n - 2] - 3d [n - 3] + 2d [n - 4] - 6d [n - 5] (C) has Z-transform 1 - 3z - 1 + 2z - 2 - 6z - 3 (D) does not satisfy any of the above three YEAR 2006

EE SP 2.59

ONE MARK

The following is true (A) A finite signal is always bounded (B) A bounded signal always possesses finite energy (C) A bounded signal is always zero outside the interval [- t0, t0] for some t0 (D) A bounded signal is always finite


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Signals and Systems

Page 97

x (t) is a real valued function of a real variable with period T . Its trigonometric Fourier Series expansion contains no terms of frequency w = 2p (2k) /T; k = 1, 2g Also, no sine terms are present. Then x (t) satisfies the equation (A) x (t) =- x (t - T) (B) x (t) = x (T - t) =- x (- t) (C) x (t) = x (T - t) =- x (t - T/2) (D) x (t) = x (t - T) = x (t - T/2)

EE SP 2.61

A discrete real all pass system has a pole at z = 2+30% : it, therefore (A) also has a pole at 12 +30% (B) has a constant phase response over the z -plane: arg H (z) = constant constant

nodia (C) is stable only if it is anti-causal

(D) has a constant phase response over the unit circle: arg H (eiW) = constant

YEAR 2006

EE SP 2.62

EE SP 2.63

TWO MARKS

x [n] = 0; n < - 1, n > 0, x [- 1] =- 1, x [0] = 2 is the input and y [n] = 0; n < - 1, n > 2, y [- 1] =- 1 = y [1], y [0] = 3, y [2] =- 2 is the output of a discrete-time LTI system. The system impulse response h [n] will be (A) h [n] = 0; n < 0, n > 2, h [0] = 1, h [1] = h [2] =- 1 (B) h [n] = 0; n < - 1, n > 1, h [- 1] = 1, h [0] = h [1] = 2 (C) h [n] = 0; n < 0, n > 3, h [0] =- 1, h [1] = 2, h [2] = 1 (D) h [n] = 0; n < - 2, n > 1, h [- 2] = h [1] = h [- 1] =- h [0] = 3 n The discrete-time signal x [n] X (z) = n3= 0 3 z2n , where 2+n transform-pair relationship, is orthogonal to the signal n (A) y1 [n] ) Y1 (z) = n3= 0 ` 2 j z - n 3

/

denotes a

/ (B) y2 [n] ) Y2 (z) = / n3= 0 (5n - n) z - (2n + 1) (C) y3 [n] ) Y3 (z) = / n3=- 3 2 - n z - n (D) y4 [n] ) Y4 (z) = 2z - 4 + 3z - 2 + 1 EE SP 2.64

A continuous-time system is described by y (t) = e - x (t) , where y (t) is the output and x (t) is the input. y (t) is bounded (A) only when x (t) is bounded (B) only when x (t) is non-negative (C) only for t # 0 if x (t) is bounded for t $ 0 (D) even when x (t) is not bounded

EE SP 2.65

The running integration, given by y (t) =

t

#- 3 x (t') dt'

(A) has no finite singularities in its double sided Laplace Transform Y (s) (B) produces a bounded output for every causal bounded input


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Signals and Systems

Chapter 2

(C) produces a bounded output for every anticausal bounded input (D) has no finite zeroes in its double sided Laplace Transform Y (s) YEAR 2005 EE SP 2.66

For the triangular wave from shown in the figure, the RMS value of the voltage is equal to

nodia (A)

1 6

(C) 1 3

EE SP 2.67

TWO MARKS

(B)

(D)

1 3 2 3

2 The Laplace transform of a function f (t) is F (s) = 5s 2+ 23s + 6 as t " 3, f (t) s (s + 2s + 2) approaches (A) 3 (B) 5 (D) 3 (C) 17 2

EE SP 2.68

The Fourier series for the function f (x) = sin2 x is (A) sin x + sin 2x (B) 1 - cos 2x (C) sin 2x + cos 2x (D) 0.5 - 0.5 cos 2x

EE SP 2.69

If u (t) is the unit step and d (t) is the unit impulse function, the inverse z -transform of F (z) = z +1 1 for k > 0 is (A) (- 1) k d (k)

(B) d (k) - (- 1) k

(C) (- 1) k u (k)

(D) u (k) - (- 1) k

YEAR 2004

TWO MARKS

EE SP 2.70

The rms value of the resultant current in a wire which carries a dc current of 10 A and a sinusoidal alternating current of peak value 20 is (A) 14.1 A (B) 17.3 A (C) 22.4 A (D) 30.0 A

EE SP 2.71

The rms value of the periodic waveform given in figure is


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Signals and Systems

Page 99

(A) 2 6 A

(B) 6 2 A

(C)

(D) 1.5 A

4/3 A

***********

nodia


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Signals and Systems

Chapter 2

SOLUTION SOL 2.1

Correct option is (B). Given the Laplace transform, X ^s h =

3s + 5 s2 + 10s + 21

From final value theorem, we know

x ^ t h = lim sX ^s h s"3

So, we obtain

x ^0+h = lim sX ^0+h

nodia s"3

= lim s d s"3

= lim s s"3

=

SOL 2.2

3s + 5 n s2 + 10s + 21

s _3 + s5 i s 2 `1 + 10s + 21 s j 2

3+ 5 3 =3 1 + 10 + 21 3 3

Correct option is (C). Consider a periodic square wave shown below.

For the square wave, we observe that the Fourier series coefficients exist and the reconstruction converges at most point. SOL 2.3

Correct option is (B). Given the Fourier transform pair, F ^w h =

3

# f ^t he

-jwt

dt

and the function g ^ t h is defined as -3

g^t h =

3

# Fû he

-jut

du

...(1)

-3

Again, the inverse Fourier transform of F ^wh is written as f ^t h =

3

# F ^w h e

jwt

dw

-3

Replacing t by - t and w by u in above expression, we get f ^- t h =

3

# Fû he

-jut

du

-3

...(2)


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Signals and Systems

Page 101

Thus, by comparing equations (1) and (2), we have g ^ t h = f ^- t h If f ^ t h be an even function, then f ^- t h = f ^ t h which in turn gives g (t ) = f (t ) i.e. g (t ) will be proportional to f ^ t h, if f ^ t h is an even function. SOL 2.4

Correct option is (A). We have the waveform,

nodia

The function represented in the waveform can be resolved in the following four waveforms; where u (t) is the unit step function.

Thus, the function shown in the waveform is ^t - T h ^t - 2T h u ^t - 2T h u ^t - T h f ^ t h = u ^ t h - u ^t - T h + T T SOL 2.5

Correct option is (B). Let input

x ^ t h = k cos 3t , k = 1


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Signals and Systems

Chapter 2

y ^ t h = A sin ^3t + ah We can see that output differs only in phase. In Laplace domain Y ^s h = X ^s h H ^s h Y ^s h = X ^s h H ^s h

Y ^ jwh = X ^ jwh H ^ jwh 1 A = 1# w w2 + 16 1 A = 1# 3 9 + 16 = 1 15

For w = 3 ,

SOL 2.6

Correct option is (B).

nodia Impulse response

Laplace transform,

h ^ t h = e-5t u ^ t h H ^s h = 1 s+5

y ^ t h = e-3t u ^ t h - e-5t u ^ t h Laplace transform, Y (s) = 1 + 1 s+3 s+5 2 = ^s + 3h^s + 5h We know that, Y (s) = X ^s h H ^s h Y ^s h So, X ^s h = H ^s h Substituting Y ^s h and H ^s h, we get X ^s h = 2 s+3 Taking inverse Laplace, transform Output,

x ^ t h = 2e-3t u ^ t h

SOL 2.7

Correct option is (A). Characteristic equation zI - A = 0 a a-1 A => a+1 a H

where

z zI - A = > 0

So,

z a a-1 H-> H 0 a+1 a

z - a -a + 1 H => -a - 1 z - a For a = 1, Characteristic equation

z-1 0 H zI - A = > -2 z - 1 zI - A = 0 ^z - 1h2 = 0

z = 1 ! j0 The roots of characteristic equation gives the system poles.


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Signals and Systems

Correct option is (C). Input signal, Sample signal,

Page 103

x ^ t h = 2 + 5 sin ^100pt h x ^nTs h = 2 + 5 sin ^100p # nTs h Ts " Sampling period, Ts = 1 400 x ^nTs h = 2 + 5 sin b100p # n # 1 l 400 x ^nTs h = 2 + 5 sin a np k 4

Transfer function

H ^z h =

N Y ^z h = 1 c 1 - z-1 m N 1-z X ^z h N-1

z-n = 1 N n=0 1 0 # n # N-1 N =* 0 otherwise 400 = =8 f = 100p = 50 Hz 50 2p 1 0#n#7 = *8 0 otherwise = x ^n h * h ^n h 5 5 = )2, 2 + 2 , ^2 + 5h, c 2 + 2 m, 2 .......3

/

nodia h ^n h

or

N=

fs f

h ^n h

So

y ^n h x ^n h

Solving the convolution

y ^n h =

N-1

/ x^k hh^n - k h

n=0

The integration sum of sine term will be zero and for constant term y ^n h =

7

/ 2 # 18 = 2

n=0

SOL 2.9

Correct option is (A). We have the waveform of function f (t) as

From the waveform, we observe that f (t) is an odd function, i.e. f (- t) =- f (t) Hence, the Fourier transform of the function is imaginary. SOL 2.10

Correct option is (A). Given the signal,


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Signals and Systems

Chapter 2

1, t < 1 x (t ) = * 0, t > 1 The waveform for the given signal is drawn as

Since, the waveform of rectangular function rect (t ) is

nodia So, comparing the two waveforms, we get

x (t ) = rect a t k 2 Now, the Fourier transform pair for rectangular function is rect c t m * t sin C c wt m t 2p so, Similarly,

rect b t l * 2 sin C c w m = 2 sin w 2 p w rect c t m * 4 sin C b 2w l = p 4

4 sin ^2wh 2 sin ^2wh = 2w w

sin ^ w2 h w rect (t ) * sin C c m = w 2p 2

and

Therefore, we have the fourier transforms F "x (2t), = F "rect (t ), sin ^ w2 h = w 2

2 sin (2w) w Thus, we obtain the fourier transform of signal y (t ) as F {y (t )} = F {x (2t) * x (t/2)} = F {x (2t)} .F {x (t/2)} sin ^ w2 h 2 sin (2w) = w w 2 and

F {x (t/2)} = F {rect (t/4)} =

= 42 sin b w l sin (2w) 2 w SOL 2.11

Correct answer is 14 Hz. Given the signal,


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Signals and Systems

Page 105

f ^ t h = 3 sin 8pt + 6 sin 12pt + sin 14pt According to Nyquist criterion the sampling frequency is f s > 2f max where fmax is the maximum frequency of the message signal. For the given signal, we have maximum frequency as or

w max = 14p 2pf max = 14p

or f max = 7 Thus, the sampling frequency is fs > 2 # 7 or fs > 14 i.e. minimum sampling frequency is 14 Hz.

SOL 2.12

nodia

Correct option is (D). To obtain the system function, we redraw the given pole-zero plot as

From the pole-zero plot, we obtain the transfer function as (S - Z1) (S - Z 1*) (S - Z2) (S - Z 2*) H ^w h = K (S - P1) (S - P 1*) (S - P2) (S - P 2*) (S - Z1) (S - Z 1*) (S - Z2) (S - Z 2*) So, | H (w) | = K (S - P1) (S - P 1*) (S - P2) (S - P 2*) Since, we can observe from the pole-zero plot that

...(1)

| P1 | = | Z2 | and | P2 | = | Z1 | Thus, the numerator and denominator terms of equation are cancelled. Hence, we get | H (w) | = K SOL 2.13

Correct option is (C). Given the period of sampling train, TS = 20 ms 1 So, the sampling frequency is fS = = 50 Hz 20 # 10-3 Let the frequency of signal x (t ) be fx After sampling the signal, the sampled signal has the frequency fS - fx = 50 - fx and fS + fx = 50 + fx Now, the sampled signal is applied to an ideal low-pass filter with cut off frequency, fc = 25 Hz Since, the output of the filter carried a single frequency component of 20 Hz. So,


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Signals and Systems

Chapter 2

it is clear that only lower component (fs - fx ) passes through the filter, i.e. fS - fx < 25 and fS - fx = 20 50 - fx = 20 fx = 50 - 20 = 30 Hz SOL 2.14

Correct option is (B). Given the relation, y (t ) = d x (t ) dt From the properties of Fourier transform we know that if y (t ) = d x (t ) dt

nodia

...(1) Y (w) = jwX (w) where X (w) and Y (w) are the Fourier transform of x (t ) and y (t ) respectively. Now, we have the function x (t ), which is differentiable, non-constant, even Then

function. So, its frequency response will be real, i.e. X (w) is real Hence, equation (1) gives the result that Y (w) is imaginary.

SOL 2.15

Correct option is (A). Given, the maximum frequency of the band-limited signal

fm = 5 kHz According to the Nyquist sampling theorem, the sampling frequency must be greater than the Nyquist frequency which is given as fN = 2fm = 2 # 5 = 10 kHz So, the sampling frequency fs must satisfy

fs $ fN fs $ 10 kHz only the option (A) does not satisfy the condition therefore, 5 kHz is not a valid sampling frequency. SOL 2.16

Correct option is (A). Given, the signal v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p4 h So we have w1 = 100 rad/s ; w2 = 300 rad/s and w3 = 500 rad/s Therefore, the respective time periods are T1 = 2p = 2p sec w1 100 T2 = 2p = 2p sec w2 300 T3 = 2p sec 500 So, the fundamental time period of the signal is LCM ^2p, 2p, 2ph L.C.M. ^T1, T2 T3h = HCF ^100, 300, 500h


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Signals and Systems

Page 107

T0 = 2p 100

or,

Thus, the fundamental frequency in rad/sec is w0 = 2p = 100 rad/s 10 SOL 2.17

SOL 2.18

SOL 2.19

Correct option is (C). If the two systems with impulse response h1 ^ t h and h2 ^ t h are connected in cascaded configuration as shown in figure, then the overall response of the system is the convolution of the individual impulse responses.

nodia

Correct option is (C). For a system to be casual, the R.O.C of system transfer function H ^s h which is rational should be in the right half plane and to the right of the right most pole. For the stability of LTI system. All poles of the system should lie in the left half of S -plane and no repeated pole should be on imaginary axis. Hence, options (A), (B), (D) satisfies both stability and causality an LTI system. But, Option (C) is not true for the stable system as, S = 1 have one pole in right hand plane also. Correct option is (C). Given, the input x ^ t h = u ^t - 1h It’s Laplace transform is -s X ^s h = e s The impulse response of system is given

h^t h = t u^t h Its Laplace transform is H ^s h = 12 s Hence, the overall response at the output is -s Y ^s h = X ^s h H ^s h = e 3 s Its inverse Laplace transform is ^t - 1h2 y^t h = u ^t - 1h 2 SOL 2.20

Correct option is (B). Given, the impulse response of continuous time system h ^ t h = d ^t - 1h + d ^t - 3h


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Signals and Systems

Chapter 2

From the convolution property, we know x ^ t h * d ^t - t 0h = x ^t - t 0h So, for the input

x ^ t h = u ^ t h (Unit step fun n ) The output of the system is obtained as y^t h = u^t h * h^t h = u ^ t h * 6d ^t - 1h + d ^t - 3h@ = u ^t - 1h + u ^t - 3h

y ^2 h = u ^2 - 1h + u ^2 - 3h = 1

At t = 2 SOL 2.21

Correct option is (C). n n x [ n] = b 1 l - b 1 l u [ n] 3 2 n -n n = b 1 l u [n] + b 1 l u [- n - 1] - b 1 l u (n) 3 3 2 Taking z -transform

nodia X 6z @ = -

=

3

3

1 n -n 1 -n -n b 3 l z u [n ] + b 3 l z u [- n - 1] n =- 3 n =- 3

/

/

3

3

-1

3

1 n -n 1 n -n 1 -n -n 1 n -n b 2 l z u [n ] = b3l z + b3l z b2l z n =- 3 n=0 n =- 3 n=0

/ 3

/

3

/ b 31z l + / b 13 z l

n=0

n

14 42 4 43 I

m

-

m=1

1 44 2 44 3 II

/

3

/ b 21z l

n

/

Taking m =- n

n=0

14 42 4 43 III

1 < 1 or z > 1 3 3z Series II converges if 1 z < 1 or z < 3 3 Series III converges if 1 < 1 or z > 1 2z 2 Region of convergence of X (z) will be intersection of above three So, ROC : 1 < z < 3 2

Series I converges if

SOL 2.22

Correct option is (D). Using s -domain differentiation property of Laplace transform. If

f (t)

F (s) dF (s) L tf (t) ds L [tf (t)] = - d ; 2 1 ds s + s + 1E = 2 2s + 1 2 (s + s + 1)

So,

SOL 2.23

L

Correct option is (A). Convolution sum is defined as y [n] = h [n] * g [n] = For causal sequence,

y [n] =

3

/ h [n] g [n - k] k =- 3

3

/ h [n] g [n - k] k=0


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Signals and Systems

Page 109

y [n] = h [n] g [n] + h [n] g [n - 1] + h [n] g [n - 2] + ..... For n = 0 ,

y [0] = h [0] g [0] + h [1] g [- 1] + ........... = h [0] g [0] = h [0] g [0]

For n = 1,

g [- 1] = g [- 2] = ....0 ...(i)

y [1] = h [1] g [1] + h [1] g [0] + h [1] g [- 1] + .... = h [1] g [1] + h [1] g [0] 1 = 1 g [1] + 1 g [0] 2 2 2

1 h [1] = b 1 l = 1 2 2

1 = g [1] + g [0] From equation (i),

nodia g [1] = 1 - 1 = 0

So,

SOL 2.24

g [1] = 1 - g [0] y [0] 1 = =1 g [0] = h [0] 1


(2 cos w) (sin 2w) = sin 3w + sin w w w w We know that inverse Fourier transform of sin c function is a rectangular function. H (jw) =

So, inverse Fourier transform of H (jw) h (t) = h1 (t) + h2 (t) h (0) = h1 (0) + h2 (0) =1+1 =1 2 2 SOL 2.25

Correct option is (D). y (t) = Time invariance : Let,

t

# x (t) cos (3t) dt -3

x (t) = d (t) y (t) =

t

# d (t) cos (3t) dt -3


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Signals and Systems

Chapter 2

= u (t) cos (0) = u (t) For a delayed input (t - t 0) output is y (t, t 0) =

t

# d (t - t ) cos (3t) dt 0

-3

= u (t) cos (3t 0) y (t - t 0) = u (t - t 0)

Delayed output

y (t, t 0) ! y (t - t 0) System is not time invariant. Stability : Consider a bounded input x (t) = cos 3t y (t) =

#

t

-3

cos2 3t =

t = 1 1dt - 1 2 -3 2 As t " 3, y (t) " 3 (unbounded) System is not stable.

#

SOL 2.26

1 - cos 6t 2 -3

#

t

t

# cos 6t dt

nodia -3

Correct option is (D).

f (t) = a 0 +

3

/ (an cos wt + bn sin nwt)

n=1

The given function f (t) is an even function, therefore bn = 0 f (t) is a non zero average value function, so it will have a non-zero value of a 0 T/2 a 0 = 1 # f (t) dt (average value of f (t)) ^T/2h 0 • an is zero for all even values of n and non zero for odd n T an = 2 # f (t) cos (nwt) d (wt) T 0

• •

So, Fourier expansion of f (t) will have a 0 and an , n = 1, 3, 5f3 SOL 2.27

Correct option is (A). x (t) = e-t Laplace transformation X (s) =

1 s+1

y (t) = e-2t Y (s) = 1 s+2 Convolution in time domain is equivalent to multiplication in frequency domain. z (t) = x (t) ) y (t)


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Signals and Systems

Page 111

Z (s) = X (s) Y (s) = b 1 lb 1 l s+1 s+2 By partial fraction and taking inverse Laplace transformation, we get Z (s) = 1 - 1 s+1 s+2 z (t) = e-t - e-2t SOL 2.28

Correct option is (D). L

f (t)

F1 (s)

e-st F1 (s) = F2 (s) F (s) F 1)(s) e-st F1 (s) F 1)(s) = G (s) = 2 F1 (s) 2 F1 (s) 2 e-sE F1 (s) 2 ) 2 = "a F1 (s) F 1 (s) = F1 (s) F1 (s) 2 = e-st Taking inverse Laplace transform L

f (t - t)

nodia g (t) = L - 1 [e-st] = d (t - t)

SOL 2.29


h (t) = e-t + e-2t

Laplace transform of h (t) i.e. the transfer function H (s) = 1 + 1 s+1 s+2 For unit step input r (t) = m (t) R (s) = 1 s

or

Y (s) = R (s) H (s)

Output,

= 1; 1 + 1 E s s+1 s+2

By partial fraction Y (s) = 3 - 1 - b 1 l 1 2s s + 1 s+2 2 Taking inverse Laplace e-2t u (t) y (t) = 3 u (t) - e-t u (t) 2 2 = u (t) 61.5 - e-t - 0.5e-2t@ SOL 2.30

Correct option is (C). System is given as 2 (s + 1) R (s) = 1 s

H (s) = Step input Output

Y (s) = H (s) R (s) = 2 b1l (s + 1) s


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Signals and Systems

Chapter 2

=2- 2 s (s + 1) Taking inverse Laplace transform y (t) = (2 - 2e- t) u (t) Final value of y (t), yss (t) = lim y (t) = 2 t"3

Let time taken for step response to reach 98% of its final value is ts . So, 2 - 2e- ts = 2 # 0.98 0.02 = e- ts ts = ln 50 = 3.91 sec. SOL 2.31

nodia Correct option is (D). Period of x (t),

T = 2p = 2 p 0.8 p w = 2.5 sec

SOL 2.32

Correct option is (B). Input output relationship y (t) =

5t

#- 3x (t) dt,

t>0

Causality : y (t) depends on x (5t), t > 0 system is non-causal. For example t = 2 y (2) depends on x (10) (future value of input) Linearity : Output is integration of input which is a linear function, so system is linear. SOL 2.33

Correct option is (A). Fourier series of given function x (t) = A0 +

3

/

an cos nw0 t + bn sin nw0 t

n=1

So,

a x (t) =- x (t) odd function A0 = 0 an = 0 T bn = 2 x (t) sin nw0 t dt T 0

#

T/2 T = 2= (1) sin nw0 t dt + (- 1) sin nw0 t dt G T 0 T/2 T/2 T = 2 =c cos nw0 t m - c cos nw0 t m G - nw0 0 - nw0 T/2 T = 2 6(1 - cos np) + (cos 2np - cos np)@ nw0 T = 2 61 - (- 1) n @ np

#

#


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Signals and Systems

Page 113

4 , n odd bn = * np 0 , n even So only odd harmonic will be present in x (t) For second harmonic component (n = 2) amplitude is zero. SOL 2.34

Correct option is (D). By parsval’s theorem 1 3 X (w) 2 dw = 2p - 3

#

#- 3

3

SOL 2.35

#- 3 x2 (t) dt 3

X (w) 2 dw = 2p # 2 = 4p

nodia Correct option is (C). Given sequences

x [n] = {1, - 1}, 0 # n # 1 -

y [n] = {1, 0, 0, 0, - 1}, 0 # n # 4 If impulse response is h [n] then y [ n ] = h [ n ] * x [ n] Length of convolution (y [n]) is 0 to 4, x [n] is of length 0 to 1 so length of h [n] will be 0 to 3. Let h [n] = {a, b, c, d} Convolution

y [n] = {a, - a + b, - b + c, - c + d, - d} Comparing we have a =1 -a + b = 0 & b = a = 1 -b + c = 0 & c = b = 1 -c + d = 0 & d = c = 1 So,

SOL 2.36

h [n] = {1, 1, 1, 1} -

Correct option is (D). We can observe that if we scale f (t) by a factor of 1 and then shift, we will get 2 g (t). First scale f (t) by a factor of 1 2 g1 (t) = f (t/2)


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Signals and Systems

Chapter 2

g (t) = g1 (t - 3) = f` t - 3 j 2

Shift g1 (t) by 3,

nodia g (t) = f` t - 3 j 2 2

SOL 2.37

Correct option is (C). g (t) can be expressed as

g (t) = u (t - 3) - u (t - 5) By shifting property we can write Laplace transform of g (t) G (s) = 1 e - 3s - 1 e - 5s s s - 3s = e (1 - e - 2s) s

SOL 2.38

Correct option is (D). L Let X (s) x (t) L Y (s) y (t) L H (s) h (t) So output of the system is given as Y (s) = X (s) H (s) Now for input So now output is

x (t - t)

L

e - st X (s)

h (t - t)

L

e- st H (s)

(shifting property)

Y' (s) = e - st X (s) $ e - ts H (s) = e - 2st X (s) H (s) = e - 2st Y (s) y' (t) = y (t - 2t)

SOL 2.39

Correct option is (B). Let three LTI systems having response H1 (z), H2 (z) and H 3 (z) are Cascaded as showing below

Assume

H1 (z) = z2 + z1 + 1 (non-causal)


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Signals and Systems

Page 115

H2 (z) = z3 + z2 + 1 (non-causal) Overall response of the system H (z) = H1 (z) H2 (z) H3 (z) H (z) = (z2 + z1 + 1) (z3 + z2 + 1) H3 (z) To make H (z) causal we have to take H3 (z) also causal. Let

H3 (z) = z - 6 + z - 4 + 1 = (z2 + z1 + 1) (z3 + z2 + 1) (z - 6 + z - 4 + 1) H (z) " causal

Similarly to make H (z) unstable atleast one of the system should be unstable. SOL 2.40

Correct option is (C). Given signal

nodia x (t) =

3

/ ak e j2pkt/T

k =- 3

Let w0 is the fundamental frequency of signal x (t) x (t) =

3

/ ak e jkw t 0

k =- 3

a 2p = w0 T

x (t) = a - 2 e - j2w t + a - 1 e - jw t + a0 + a1 e jw t + a2 e j2w t 0

0

0

0

= (2 - j) e - 2jw t + (0.5 + 0.2j) e - jw t + 2j + 0

0

+ (0.5 - 0.2) e jw t + (2 + j) e j2w t 0

0

= 2 6e - j2w t + e j2w t @ + j 6e j2w t - e - j2w t @ + 0

0

0

0

0.5 6e jw t + e - jw t @ - 0.2j 6e+ jw t - e - jw t @ + 2j = 2 (2 cos 2w0 t) + j (2j sin 2w0 t) + 0.5 (2 cos w0 t) 0.2j (2j sin w0 t) + 2j = 64 cos 2w0 t - 2 sin 2w0 t + cos w0 t + 0.4 sin w0 t @ + 2j 0

0

0

0

Im [x (t)] = 2 (constant) SOL 2.41

Correct option is (A). Z-transform of x [n] is X (z) = 4z - 3 + 3z - 1 + 2 - 6z2 + 2z3 Transfer function of the system H (z) = 3z - 1 - 2 Output Y (z) = H (z) X (z) Y (z) = (3z - 1 - 2) (4z - 3 + 3z - 1 + 2 - 6z2 + 2z3) = 12z -4 + 9z -2 + 6z -1 - 18z + 6z2 - 8z -3 - 6z -1 - 4 + 12z2 - 4z3 = 12z - 4 - 8z - 3 + 9z - 2 - 4 - 18z + 18z2 - 4z3 Or sequence y [n] is y [n] = 12d [n - 4] - 8d [n - 3] + 9d [n - 2] - 4d [n] 18d [n + 1] + 18d [n + 2] - 4d [n + 3] y [n] = Y 0, n < 0 So y [n] is non-causal with finite support.

SOL 2.42



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Signals and Systems

Chapter 2

Since the given system is LTI, So principal of Superposition holds due to linearity. For causal system h (t) = 0 , t < 0 Both statement are correct. SOL 2.43

Correct option is (C). For an LTI system output is a constant multiplicative of input with same frequency. Here

input g (t) = e - at sin (wt)

output y (t) = Ke - bt sin (vt + f) Output will be in form of Ke - at sin (wt + f) So \= b, v = w SOL 2.44

nodia Correct option is (D).

z , z >a (z - a) 2 at z = a is = d (z - a) 2 X (z) zn - 1 z = a dz z = d (z - a) 2 zn - 1 dz (z - a) 2 z=a n d = z dz z = a X (z) =

Given that

Residue of X (z) zn - 1

= nzn - 1 z = a = nan - 1

SOL 2.45

Correct option is (D). Input-output relation y (t) =

- 2t

#- 3x (t) dt

Causality : Since y (t) depends on x (- 2t), So it is non-causal. Time-variance : y (t) =

- 2t

#- 3x (t - t0) dt =Y y (t - t0)

So this is time-variant. Stability : Output y (t) is unbounded for an bounded input. For example Let

x (t) = e - t (bounded) y (t) =

SOL 2.46

- 2t

- t - 2t

#- 3e- t dt = 8-e 1 B- 3 $ Unbounded

Correct option is (A). Output y (t) of the given system is Or Given that,

y (t) = x (t) ) h (t) Y (jw) = X (jw) H (jw) x (t) = sinc (at) and h (t) = sinc (bt)


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Signals and Systems

Page 117

Fourier transform of x (t) and h (t) are X (jw) = F [x (t)] = p rect` w j, - a < w < a a 2a H (jw) = F [h (t)] = p rect` w j, - b < w < b b 2b 2 Y (jw) = p rect` w j rect` w j ab 2a 2b

nodia Y (jw) = K rect ` w j 2g

So,

g = min (a, b) y (t) = K sinc (gt)

Where And

SOL 2.47

Correct option is (B). Let ak is the Fourier series coefficient of signal x (t) Given y (t) = x (t - t0) + x (t + t0) Fourier series coefficient of y (t) bk = e - jkwt ak + e jkwt ak bk = 2ak cos kwt0 bk = 0 (for all odd k ) kwt0 = p , k " odd 2 k 2p t0 = p 2 T t0 = T 4 0

For k = 1, SOL 2.48


SOL 2.49

Correct option is (C). Given signal

So, Similarly

0

x (t) = rect `t - 1 j 2 1, - 1 # t - 1 # 1 or 0 # t # 1 2 2 2 x (t) = * 0, elsewhere x (- t) = rect`- t - 1 j 2 1, - 1 # - t - 1 # 1 or - 1 # t # 0 2 2 2 x (- t) = * 0, elsewhere


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Signals and Systems

F [x (t) + x (- t)] = =

Chapter 2

#- 3 x (t) e- jwt dt + #- 3 x (- t) e- jwt dt 3

3

0

1

#0 (1) e- jwt dt + #- 1 (1) e- jwt dt 1

0

- jw t - jw t = ;e E +;e E - jw 0 - jw - 1 = 1 (1 - e - jw) + 1 (e jw - 1) jw jw - jw /2 jw /2 =e (e jw/2 - e - jw/2) + e (e jw/2 - e - jw/2) jw jw

(e jw/2 - e - jw/2) (e - jw/2 + e jw/2) jw = 2 sin ` w j $ 2 cos ` w j w 2 2 = 2 cos w sinc` w j 2 2p

=

SOL 2.50

nodia Correct option is (B). In option (A)

z1 [n] = x [n - 3] z2 [n] = z1 [4n] = x [4n - 3] y [n] = z2 [- n]

= x [- 4n - 3] = Y x [3 - 4n]

In option (B)

z1 [n] = x [n + 3] z2 [n] = z1 [4n] = x [4n + 3] y [n] = z2 [- n] = x [- 4n + 3] In option (C) v1 [n] = x [4n] v2 [n] = v1 [- n] = x [- 4n] y [n] = v2 [n + 3] = x [- 4 (n + 3)] = Y x [3 - 4n]

In option (D) v1 [n] = x [4n] v2 [n] = v1 [- n] = x [- 4n] y [n] = v2 [n - 3] = x [- 4 (n - 3)] = Y x [3 - 4n] SOL 2.51

Correct option is (D). For an LTI system input and output have identical wave shape (i.e. frequency of input-output is same) within a multiplicative constant (i.e. Amplitude response is constant) So F must be a sine or cosine wave with w1 = w2

SOL 2.52

Correct option is (D). The spectrum of sampled signal s (jw) contains replicas of U (jw) at frequencies !nfs . Where n = 0, 1, 2.......


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Signals and Systems

Page 119

1 = 1 kHz fs = 1 = Ts 1 m sec

nodia SOL 2.53

Correct option is (C). Given signal has the following wave-form

Function x(t) is periodic with period 2T and given that x (t) =- x (t + T) (Half-wave symmetric) So we can obtain the Fourier series representation of given function. SOL 2.54

Correct option is (A). System response is given as


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Signals and Systems

Chapter 2

G (z) H (z) = 1 - KG (z) g [n] = d [n - 1] + d [n - 2] G (z) = z - 1 + z - 2 (z - 1 + z - 2) = 2 z+1 -1 -2 z - Kz - K 1 - K (z + z ) For system to be stable poles should lie inside unit circle. So

H (z) =

z #1 2 z = K ! K + 4K # 1 K ! K2 + 4K # 2 2

K 2 + 4K # 2 - K K2 + 4K # 4 - 4K + K2

nodia 8K # 4

K # 1/2

SOL 2.55

Correct option is (C). Output is said to be distortion less if the input and output have identical wave shapes within a multiplicative constant. A delayed output that retains input waveform is also considered distortion less. Thus for distortion less output, input-output relationship is given as y (t) = Kg (t - td ) Taking Fourier transform.

Y (w) = KG (w) e - jwt = G (w) H (w) H (w) & transfer function of the system d

H (w) = Ke - jwt Amplitude response H (w) = K Phase response, qn (w) =- wtd For distortion less output, phase response should be proportional to frequency. So,

SOL 2.56

d

Correct option is (A). G (z) z = e = ae- jw + be- 3jw for linear phase characteristic a = b . jw

SOL 2.57

Correct option is (C). Given Convolution is, h (t) = u (t + 1) ) r (t - 2) Taking Laplace transform on both sides, H (s) = L [h (t)] = L [u (t + 1)] ) L [r (t - 2)] We know that, L [u (t)] = 1/s

and

So

(Time-shifting property) L [u (t + 1)] = es c 12 m s L [r (t)] = 1/s2 (Time-shifting property) L [r (t - 2) = e - 2s c 12 m s H (s) = ;es ` 1 jE;e - 2s c 12 mE s s


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Signals and Systems

Page 121

H (s) = e - s c 13 m s Taking inverse Laplace transform h (t) = 1 (t - 1) 2 u (t - 1) 2 SOL 2.58

Correct option is (C). Impulse response of given LTI system. h [ n ] = x [ n - 1] ) y [ n ] Taking z -transform on both sides. H (z) = z - 1 X (z) Y (z) We have X (z) = 1 - 3z - 1 and Y (z) = 1 + 2z - 2 So

a x [n - 1]

nodia

Z

z - 1 x (z)

H (z) = z - 1 (1 - 3z - 1) (1 + 2z - 2) Output of the system for input u [n] = d [n - 1] is , y (z) = H (z) U (z)

U [n ]

Z

U (z) = z - 1

So

Y (z) = z - 1 (1 - 3z - 1) (1 + 2z - 2) z - 1

= z - 2 (1 - 3z - 1 + 2z - 2 - 6z - 3)

= z - 2 - 3z - 3 + 2z - 4 - 6z - 5 Taking inverse z-transform on both sides we have output.

y [n] = d [n - 2] - 3d [n - 3] + 2d [n - 4] - 6d [n - 5]

SOL 2.59

Correct option is (B). A bounded signal always possesses some finite energy. t0

#- t

E =

g (t) 2 dt < 3

0

SOL 2.60

Correct option is (C). Trigonometric Fourier series is given as 3

/ an cos nw0 t + bn sin nw0 t

x (t) = A0 +

n=1

Since there are no sine terms, so bn = 0 bn = 2 T0

#0

= 2= T0

T0

#0

x (t) sin nw0 t dt

T0 /2

x (t) sin nw0 t dt +

T

#T /2 x (t) sin nw0 t dt G 0

Where t = T - t & dt =- dt = 2; T0 = 2; T0 = 2; T0

T0 /2

#T

0

T

#T /2 x (t) sin nw0 t dt E 0

TO

T

#T /2 x (T - t) sin n` 2Tp T - t j dt ++ #T /2 x (t) sin nw0 t dt E O

0

T0

T0

#T /2 x (T - t) sin (2np - nw0) dt+ #T /2 x (t) sin nw0 t dt E

= 2 ;T0 bn = 0 if

x (T - t) sin nw0 (T - t) (- dt)+

0

0

T0

T0

0

0

#T /2 x (T - t) sin (nw0 t) dt + + #T /2 x (t) sin nw0 t dt E

x (t) = x (T - t)


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Signals and Systems

Chapter 2

From half wave symmetry we know that if x (t) =- x`t ! T j 2 Then Fourier series of x (t) contains only odd harmonics. SOL 2.61

Correct option is (C). Z -transform of a discrete all pass system is given as -1 ) H (z) = z - z-0 1 1 - z0 z It has a pole at z 0 and a zero at 1/z) 0. Given system has a pole at z = 2+30% = 2

( 3 + j) = ( 3 + j) 2

nodia system is stable if z < 1 and for this it is anti-causal. SOL 2.62

Correct option is (A). According to given data input and output Sequences are x [n] = {- 1, 2}, - 1 # n # 0 y [n] = {- 1, 3, - 1, - 2}, - 1 # n # 2 If impulse response of system is h [n] then output y [n] = h [n] ) x [n] Since length of convolution (y [n]) is - 1 to 2, x [n] is of length - 1 to 0 so length of h [n] is 0 to 2. Let h [n] = {a, b, c} Convolution

So, a = 1

y [n] = {- a, 2a - b, 2b - c, 2c} y [n] = {- 1, 3, - 1, - 2} -


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Signals and Systems

Page 123

2a - b = 3 & b =- 1 2a - c =- 1 & c =- 1 Impulse response h [n] = "1, - 1, - 1, SOL 2.63


SOL 2.64

Correct option is (D). Output y (t) = e - x (t) If x (t) is unbounded, x (t) " 3 y (t) = e - x (t) " 0 (bounded) So y (t) is bounded even when x (t) is not bounded.

SOL 2.65

nodia Correct option is (B).

y (t) =

Given

t

# x (t') dt'

-3

Laplace transform of y (t) X (s) , has a singularity at s = 0 Y (s) = s t

# x (t') dt'

For a causal bounded input, y (t)=

is always bounded.

-3

SOL 2.66

Correct option is (A). RMS value is given by

1 T

Vrms = Where

#0

T

V2 (t) dt

2 T ` T j t, 0 # t # 2

V (t) = * So

1 T

#0

T

T
0,

V 2 (t) dt = 1 = T

#0

T/2

= 1 $ 42 T T

2t 2 ` T j dt +

#0

T /2

T

#T/2 (0) dt G

t2 dt

T/2

3 = 43 ; t E T 3 0 3 = 43 # T = 1 24 6 T

Vrms = SOL 2.67

1 V 6

Correct option is (A). By final value theorem lim f (t) = lim s F (s)

t"3

s"0

= lim s s"0

SOL 2.68

(5s2 + 23s + 6) = 6 =3 2 2 s (s + 2s + 2)



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Signals and Systems

Chapter 2

f (x) = sin2 x = 1 - cos 2x 2 = 0.5 - 0.5 cos 2x f (x) = A0 +

3

/ an cos nw0 x + bn sin nw0 x n=1

f (x) = sin2 x is an even function so bn = 0 A0 = 0.5 - 0.5, n = 1 an = ) 0 , otherwise 2 p w0 = = 2p = 2 T0 T SOL 2.69


nodia F (z) =

Z-transform

f (k) = d (k) - (- 1) k Z 1 (- 1) k 1 + z- 1

so,

Thus

SOL 2.70

1 = 1- z = 1- 1 z+1 z+1 1 + z- 1

Correct option is (B). Total current in wire

I = 10 + 20 sin wt

Irms =

SOL 2.71

(10) 2 +

(20) 2 = 17.32 A 2

Correct option is (A). Root mean square value is given as 1 T I2 (t) dt T 0 - 12 t, 0 # t < T 2 I (t) = * ` T j T/2 < t # T 6,

#

Irms =

From the graph,

So

1 T

#0

T

I2 dt = 1 = T

#0

T /2

- 12t 2 ` T j dt +

T

#T/2 (6) 2 dt G

T/2

t3 = 1 e 144 + 36 6t @TT/2 o ; T T2 3 E0 T3 + 36 T = 1 ; 144 c ` 2 jE 2 T T 24 m = 1 [6T + 18T] = 24 T Irms =

24 = 2 6 A

***********


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Electric Machines

Page 125

CHAPTER 3 ELECTRIC MACHINES

YEAR 2014 EE01

ONE MARK

EE SP 3.1

An 8-pole, 3-phase, 50 Hz induction motor is operating at a speed of 700 rpm. The frequency of the rotor current of the motor in Hz is ______.

EE SP 3.2

For a specified input voltage and frequency, if the equivalent radius of the core of a transformer is reduced by half, the factor by which the number of turns in the primary should change to maintain the same no load current is (A) 1/4 (B) 1/2

nodia (C) 2

EE SP 3.3

(D) 4

A star connected 400 V, 50 Hz, 4 pole synchronous machine gave the following open circuit and short circuit test results : Open circuit test : Voc = 400 V (rms, line-to-line) at field current, I f = 2.3 A Short circuit test : Isc = 10 A (rms, phase) at field current, I f = 1.5 A The value of per phase synchronous impedance in W at rated voltage is_____. YEAR 2014 EE01

TWO MARKS

EE SP 3.4

The core loss of a single phase, 230/115 V , 50 Hz power transformer is measured from 230 V side by feeding the primary (230 V side) from a variable voltage variable frequency source while keeping the secondary open circuited. The core loss is measured to be 1050 W for 230 V, 50 Hz input. The core loss is again measured to be 500 W for 138 V, 30 Hz input. The hysteresis and eddy current losses of the transformer for 230 V, 50 Hz input are respectively, (A) 508 W and 542 W (B) 468 W and 582 W (C) 498 W and 552 W (D) 488 W and 562 W

EE SP 3.5

A 15 kW, 230 V dc shunt motor has armature circuit resistance of 0.4 W and field circuit resistance of 230 W . At no load and rated voltage, the motor runs at 1400 rpm and the line current drawn by the motor is 5 A. At full load, the motor draws a line current of 70 A. Neglect armature reaction. The full load speed of the motor in rpm is_____.

EE SP 3.6

A 3 phase, 50 Hz, six pole induction motor has a rotor resistance of 0.1 W and reactance of 0.92 W . Neglect the voltage drop in stator and assume that the rotor resistance is constant. Given that the full load slip is 3%, the ratio of maximum torque to full load torque is (A) 1.567 (B) 1.712 (C) 1.948 (D) 2.134


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Electric Machines

Chapter 3


Assuming an ideal transformer, the Thevenin’s equivalent voltage and impedance as seen from the terminals x and y for the circuit in figure are

(A) 2 sin ^wt h, 4 W (C) 1 sin ^wt h, 2 W

EE SP 3.8

ONE MARK

(B) 1 sin ^wt h, 1 W (D) 2 sin ^wt h, 0.5 W

nodia

A single phase, 50 kVA, 1000 V/100 V two winding transformer is connected as an autotransformer as shown in the figure.

The kVA rating of the autotransformer is ________. EE SP 3.9

A three-phase, 4-pole, self excited induction generator is feeding power to a load at a frequency f 1 . If the load is partially removed, the frequency becomes f 2 . If the speed of the generator is maintained at 1500 rpm in both the cases, then (A) f 1 , f 2 > 50 Hz and f 1 > f 2 (B) f 1 < 50 Hz and f 2 > 50 Hz (C) f 1 , f 2 < 50 Hz and f 2 > f 1 (D) f 1 > 50 Hz and f 2 < 50 Hz

EE SP 3.10

A single phase induction motor draws 12 MW power at 0.6 lagging power. A capacitor is connected in parallel to the motor to improve the power factor of the combination of motor and capacitor to 0.8 lagging. Assuming that the real and reactive power drawn by the motor remains same as before, the reactive power delivered by the capacitor in MVAR is ______.

EE SP 3.11

In a constant V/f control of induction motor, the ratio V/f is maintained constant from 0 to base frequency, where V is the voltage applied to the motor at fundamental frequency f . Which of the following statements relating to low frequency operation of the motor is TRUE ? (A) At low frequency, the stator flux increases from its rated value. (B) At low frequency, the stator flux decreases from its rated value. (C) At low frequency, the motor saturates. (D) At low frequency, the stator flux remains unchanged at its rated value.


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Electric Machines

YEAR 2014 EE02

Page 127

TWO MARKS

EE SP 3.12

A 250 V dc shunt machine has armature circuit resistance of 0.6 W and field circuit resistance of 125 W . The machine is connected to 250 V supply mains. The motor is operated as a generator and then as a motor separately. The line current of the machine in both the cases is 50 A. The ratio of the speed as a generator to the speed as a motor is _____.

EE SP 3.13

A three phase slip-ring induction motor, provided with a commutator winding, is shown in the figure. The motor rotates in clockwise direction when the rotor windings are closed.

nodia

If the rotor winding is open circuited and the system is made to run at rotational speed f r with the help of prime-mover in anti-clockwise direction, then the frequency of voltage across slip rings is f 1 and frequency of voltage across commutator brushes is f 2 . The values of f 1 and f 2 respectively are (B) f - f r and f (A) f + f r and f (C) f - f r and f + f r (D) f + f r and f - f r

EE SP 3.14

A 20-pole alternators is having 180 identical stator slots with 6 conductors in each slot. All the coils of a phase are in series. If the coils are connected to realize single-phase winding, the generated voltage is V1 . If the coils are reconnected to realize three phase star connected winding, the generated phase voltage is V2 . Assuming full pitch, single-layer winding, the ratio V1 /V2 is (A) 1 (B) 1 2 3 (C) 3 (D) 2

EE SP 3.15

For a single phase, two winding transformer, the supply frequency and voltage are both increased by 10%. The percentage changes in the hysteresis loss and eddy current loss, respectively, are (A) 10 and 21 (B) - 10 and 21 (D) - 21 and 10 (C) 21 and 10 YEAR 2014 EE03

EE SP 3.16

ONE MARK

In a synchronous machine, hunting is predominantly damped by (A) mechanical losses in the rotor (B) iron losses in the rotor (C) copper losses in the stator (D) copper losses in the rotor


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Electric Machines

Chapter 3

A single phase induction motor is provided with capacitor and centrifugal switch in series with auxiliary winding. The switch is expected to operate at a speed of 0.7 Ns, but due to malfunctioning the switch fails to operate. The torque-speed characteristic of the motor is represented by

nodia EE SP 3.18

EE SP 3.19

The no-load speed of a 230 V separately excited dc motor is 1400 rpm. The armature resistance drop and the brush drop are neglected. The field current is kept constant at rated value. The torque of the motor in Nm for an armature current of 8 A is ________. The torque speed characteristics of motor ^TM h and load ^TL h for two cases are shown in the figures (a) and (b). The load torque is equal to motor torque at points P , Q , R and S .

The stable operating points are (A) P and R (C) Q and R

(B) P and S (D) Q and S


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Electric Machines


Page 129

TWO MARKS

An open circuit test is performed on 50 Hz transformer, using variable frequency source and keeping V/f ratio constant, to separate its eddy current and hysteresis losses. The variation of core loss/frequency as function of frequency is shown in the figure

nodia

The hysteresis and eddy current losses of the transformer at 25 Hz respectively are (A) 250 W and 2.5 W (B) 250 W and 62.5 W (C) 312.5 W and 62.5 W (D) 312.5 W and 250 W

EE SP 3.21

A non-salient pole synchronous generator having synchronous reactance of 0.8 pu is supplying 1 pu power to a unity power factor load at a terminal voltage of 1.1 pu. Neglecting the armature resistance, the angle of the voltage behind the synchronous reactance with respect to the angle of the terminal voltage in degrees is _______.

EE SP 3.22

A separately excited 300 V DC shunt motor under no load runs at 900 rpm drawing an armature current of 2 A. The armature resistance is 0.5 W and leakage inductance is 0.01 H. When loaded, the armature current is 15 A. Then, the speed in rpm is _____.

EE SP 3.23

The load shown in the figure absorbs 4 kW at a power factor of 0.89 lagging.

Assuming the transformer to be ideal, the value of the reactance X to improve the input power factor to unity is _________. EE SP 3.24

The parameters measured for a 220 V/110 V, 50 Hz , single phase transformer are : Self inductance of primary winding = 45 mH


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Electric Machines

Chapter 3

Self inductance of secondary winding = 30 mH Mutual inductance between primary and secondary windings = 20 mH Using the above parameters, the leakage ^Ll1, Ll2h and magnetizing ^Lm h inductance as referred to primary side in the equivalent circuit respectively, are (A) 5 mH, 20 mH and 40 mH (B) 5 mH, 80 mH and 40 mH (C) 25 mH, 10 mH and 20 mH (D) 45 mH, 30 mH and 20 mH YEAR 2013 EE SP 3.25

ONE MARK

Leakage flux in an induction motor is (A) flux that leaks through the machine

nodia (B) flux that links both stator and rotor windings (C) flux that links none of the windings

(D) flux that links the stator winding or the rotor winding but not both

EE SP 3.26

The angle d in the swing equation of a synchronous generator is the (A) angle between stator voltage and current (B) angular displacement of the rotor with respect to the stator (C) angular displacement of the stator mmf with respect to a synchronously rotating axis. (D) angular displacement of an axis fixed to the rotor with respect to a synchronously rotating axis

EE SP 3.27

A single-phase transformer has no-load loss of 64 W, as obtained from an open circuit test. When a short-circuit test is performed on it with 90% of the rated currents flowing in its both LV and HV windings, he measured loss is 81 W. The transformer has maximum efficiency when operated at (A) 50.0% of the rated current (B) 64.0% of the rated current (C) 80.0% of the rated current (D) 88.8% of the rated current YEAR 2013

TWO MARKS

EE SP 3.28

A 4-pole induction motor, supplied by a slightly unbalanced three-phase 50 Hz source, is rotating at 1440 rpm. The electrical frequency in Hz of the induced negative sequence current in the rotor is (A) 100 (B) 98 (C) 52 (D) 48

EE SP 3.29

The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage VWX1 = 100 V is applied across WX to get an open circuit voltage VYZ1 across YZ. Next, an ac voltage


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Electric Machines

Page 131

VYZ2 = 100 V is applied across YZ to get an open circuit voltage VWX2 across WX. Then, VYZ1 /VWX1 , VWX2 /VYZ2 are respectively,

(A) 125/100 and 80/100

(B) 100/100 and 80/100

(C) 100/100 and 100/100

(D) 80/100 and 80/100

nodia YEAR 2012

EE SP 3.30

ONE MARK

The slip of an induction motor normally does not depend on (A) rotor speed (B) synchronous speed (C) shaft torque

(D) core-loss component YEAR 2012

TWO MARKS

EE SP 3.31

A 220 V, 15 kW, 100 rpm shunt motor with armature resistance of 0.25 W, has a rated line current of 68 A and a rated field current of 2.2 A. The change in field flux required to obtain a speed of 1600 rpm while drawing a line current of 52.8 A and a field current of 1.8 A is (A) 18.18% increase (B) 18.18% decrease (C) 36.36% increase (D) 36.36% decrease

EE SP 3.32

The locked rotor current in a 3-phase, star connected 15 kW, 4 pole, 230 V, 50 Hz induction motor at rated conditions is 50 A. Neglecting losses and magnetizing current, the approximate locked rotor line current drawn when the motor is connected to a 236 V, 57 Hz supply is (A) 58.5 A (B) 45.0 A (C) 42.7 A (D) 55.6 A

EE SP 3.33

A single phase 10 kVA, 50 Hz transformer with 1 kV primary winding draws 0.5 A and 55 W, at rated voltage and frequency, on no load. A second transformer has a core with all its linear dimensions 2 times the corresponding dimensions of the first transformer. The core material and lamination thickness are the same in both transformer. The primary winding of both the transformers have the save number of turns. If a rate voltage of 2 kV at 50 Hz is applied to the primary of the second transformer, then the no load current and power, respectively, are (A) 0.7 A, 77.8 A (B) 0.7 A, 155.6 W


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Electric Machines

(C) 1 A, 110 W

Chapter 3

(D) 1 A, 220 W


ONE MARK

A 4 point starter is used to start and control the speed of a (A) dc shunt motor with armature resistance control (B) dc shunt motor with field weakening control (C) dc series motor (D) dc compound motor

EE SP 3.35

A three phase, salient pole synchronous motor is connected to an infinite bus. It is operated at no load a normal excitation. The field excitation of the motor is first reduced to zero and then increased in reverse direction gradually. Then the armature current. (A) Increases continuously

nodia (B) First increases and then decreases steeply

(C) First decreases and then increases steeply (D) Remains constant

EE SP 3.36

A single phase air core transformer, fed from a rated sinusoidal supply, is operating at no load. The steady state magnetizing current drawn by the transformer from the supply will have the waveform


TWO MARKS

A 220 V, DC shunt motor is operating at a speed of 1440 rpm. The armature resistance is 1.0 W and armature current is 10 A. of the excitation of the machine is reduced by 10%, the extra resistance to be put in the armature circuit to


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Electric Machines

Page 133

maintain the same speed and torque will be (B) 2.1 W (A) 1.79 W (D) 3.1 W (C) 18.9 W EE SP 3.38

A three-phase 440 V, 6 pole, 50 Hz, squirrel cage induction motor is running at a slip of 5%. The speed of stator magnetic field to rotor magnetic field and speed of rotor with respect of stator magnetic field are (A) zero, - 5 rpm (B) zero, 955 rpm (C) 1000 rpm, - 5 rpm (D) 1000 rpm, 955 rpm

nodia YEAR 2010

EE SP 3.39

A Single-phase transformer has a turns ratio 1:2, and is connected to a purely resistive load as shown in the figure. The magnetizing current drawn is 1 A, and the secondary current is 1 A. If core losses and leakage reactances are neglected, the primary current is

(A) 1.41 A (C) 2.24 A EE SP 3.40

ONE MARK

(B) 2 A (D) 3 A

A balanced three-phase voltage is applied to a star-connected induction motor, the phase to neutral voltage being V . The stator resistance, rotor resistance referred to the stator, stator leakage reactance, rotor leakage reactance referred to the stator, and the magnetizing reactance are denoted by rs , rr , Xs , Xr and Xm , respectively. The magnitude of the starting current of the motor is given by Vs Vs (A) (B) 2 2 2 (rs + rr ) + (Xs + Xr ) rs + (Xs + Xm) 2 Vs Vs (C) (D) 2 2 2 (rs + rr ) + (Xm + Xr ) rs + (Xm + Xr ) 2 YEAR 2010

EE SP 3.41

TWO MARKS

A separately excited dc machine is coupled to a 50 Hz, three-phase, 4-pole induction machine as shown in figure. The dc machine is energized first and the machines rotate at 1600 rpm. Subsequently the induction machine is also connected to a 50 Hz, three-phase source, the phase sequence being consistent with the direction of rotation. In steady state


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Electric Machines

Chapter 3

(A) both machines act as generator (B) the dc machine acts as a generator, and the induction machine acts as a motor (C) the dc machine acts as a motor, and the induction machine acts as a generator (D) both machines act as motors EE SP 3.42

nodia

A balanced star-connected and purely resistive load is connected at the secondary of a star-delta transformer as shown in figure. The line-to line voltage rating of the transformer is 110 V/200 V. Neglecting the non-idealities of the transformer, the impedance Z of the equivalent star-connected load, referred to the primary side of the transformer, is

(A) (3 + j0) W (C) (0.866 + j0.5) W

(B) (0.866 - j0.5) W (D) (1 + j0) W

Common Data for Questions 43 and 44 A separately excited DC motor runs at 1500 rpm under no-load with 200 V applied to the armature. The field voltage is maintained at its rated value. The speed of the motor, when it delivers a torque of 5 Nm, is 1400 rpm as shown in figure. The rotational losses and armature reaction are neglected.


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Electric Machines

Page 135

EE SP 3.43

The armature resistance of the motor is (A) 2 W (B) 3.4 W (C) 4.4 W (D) 7.7 W

EE SP 3.44

nodia

For the motor to deliver a torque of 2.5 Nm at 1400 rpm, the armature voltage to be applied is (A) 125.5 V (B) 193.3 V (C) 200 V (D) 241.7 V YEAR 2009

EE SP 3.45

ONE MARK

A field excitation of 20 A in a certain alternator results in an armature current of 400 A in short circuit and a terminal voltage of 2000 V on open circuit. The magnitude of the internal voltage drop within the machine at a load current of 200 A is (A) 1 V (B) 10 V (C) 100 V (D) 1000 V

EE SP 3.46

The single phase, 50 Hz iron core transformer in the circuit has both the vertical arms of cross sectional area 20 cm2 and both the horizontal arms of cross sectional area 10 cm2 . If the two windings shown were wound instead on opposite horizontal arms, the mutual inductance will

(A) double (C) be halved EE SP 3.47

(B) remain same (D) become one quarter

A 3-phase squirrel cage induction motor supplied from a balanced 3-phase source drives a mechanical load. The torque-speed characteristics of the motor(solid curve) and of the load(dotted curve) are shown. Of the two equilibrium points A and B, which of the following options correctly describes the stability of A and B ?


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Electric Machines

Chapter 3

(A) A is stable, B is unstable

(B) A is unstable, B is stable

(C) Both are stable

(D) Both are unstable


TWO MARKS

nodia

A 200 V, 50 Hz, single-phase induction motor has the following connection diagram and winding orientations as shown. MM’ is the axis of the main stator winding(M1 M2) and AA’ is that of the auxiliary winding(A1 A2). Directions of the winding axis indicate direction of flux when currents in the windings are in the directions shown. Parameters of each winding are indicated. When switch S is closed the motor

(A) (B) (C) (D)

rotates clockwise rotates anti-clockwise does not rotate rotates momentarily and comes to a halt

Common Data for Questions 49 and 50 : The circuit diagram shows a two-winding, lossless transformer with no leakage flux, excited from a current source, i (t), whose waveform is also shown. The transformer has a magnetizing inductance of 400/p mH.


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EE SP 3.49

EE SP 3.50

Electric Machines

Page 137

nodia The peak voltage across A and B, with S open is (B) 800 V (A) 400 V p (D) 800 V (C) 4000 V p p

If the wave form of i (t) is changed to i (t) = 10 sin (100pt) A, the peak voltage across A and B with S closed is (A) 400 V (B) 240 V (C) 320 V (D) 160 V

Common Data for Questions 51 and 52

The star-delta transformer shown above is excited on the star side with balanced, 4-wire, 3-phase, sinusoidal voltage supply of rated magnitude. The transformer is under no load condition EE SP 3.51

With both S1 and S2 open, the core flux waveform will be (A) a sinusoid at fundamental frequency (B) flat-topped with third harmonic (C) peaky with third-harmonic (D) none of these


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Electric Machines

Chapter 3

With S2 closed and S1 open, the current waveform in the delta winding will be (A) a sinusoid at fundamental frequency (B) flat-topped with third harmonic (C) only third-harmonic (D) none of these

EE SP 3.53

Figure shows the extended view of a 2-pole dc machine with 10 armature conductors. Normal brush positions are shown by A and B, placed at the interpolar axis. If the brushes are now shifted, in the direction of rotation, to A’ and B’ as shown, the voltage waveform VA'B' will resemble

nodia Statement for Linked Answer Questions 54 and 55

The figure above shows coils-1 and 2, with dot markings as shown, having 4000 and 6000 turns respectively. Both the coils have a rated current of 25 A. Coil-1 is excited with single phase, 400 V, 50 Hz supply. EE SP 3.54

400 The coils are to be connected to obtain a single-phase, 1000 V, auto-transformer to drive a load of 10 kVA. Which of the options given should


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Electric Machines

Page 139

be exercised to realize the required auto-transformer ? (A) Connect A and D; Common B (B) Connect B and D; Common C (C) Connect A and C; Common B (D) Connect A and C; Common D EE SP 3.55

In the autotransformer obtained in Question 16, the current in each coil is (A) Coil-1 is 25 A and Coil-2 is 10 A (B) Coil-1 is 10 A and Coil-2 is 25 A (C) Coil-1 is 10 A and Coil-2 is 15 A (D) Coil-1 is 15 A and Coil-2 is 10 A

nodia YEAR 2008

EE SP 3.56

ONE MARK

Distributed winding and short chording employed in AC machines will result in (A) increase in emf and reduction in harmonics (B) reduction in emf and increase in harmonics (C) increase in both emf and harmonics

(D) reduction in both emf and harmonics

EE SP 3.57

Three single-phase transformer are connected to form a 3-phase transformer bank. The transformers are connected in the following manner :

The transformer connecting will be represented by (A) Y d0 (B) Y d1 (C) Y d6 (D) Y d11 EE SP 3.58

In a stepper motor, the detent torque means (A) minimum of the static torque with the phase winding excited (B) maximum of the static torque with the phase winding excited (C) minimum of the static torque with the phase winding unexcited (D) maximum of the static torque with the phase winding unexcited

EE SP 3.59

It is desired to measure parameters of 230 V/115 V, 2 kVA, single-phase transformer. The following wattmeters are available in a laboratory: W1 : 250 V, 10 A, Low Power Factor W2 : 250 V, 5 A, Low Power Factor W3 : 150 V, 10 A, High Power Factor W4 : 150 V, 5 A, High Power Factor


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Electric Machines

Chapter 3

The Wattmeters used in open circuit test and short circuit test of the transformer will respectively be (A) W1 and W2 (B) W2 and W4 (C) W1 and W4

(D) W2 and W3


EE SP 3.61

A 230 V, 50 Hz, 4-pole, single-phase induction motor is rotating in the clockwise (forward) direction at a speed of 1425 rpm. If the rotor resistance at standstill is 7.8 W, then the effective rotor resistance in the backward branch of the equivalent circuit will be (A) 2 W (B) 4 W (C) 78 W (D) 156 W

nodia

A 400 V, 50 Hz 30 hp, three-phase induction motor is drawing 50 A current at 0.8 power factor lagging. The stator and rotor copper losses are 1.5 kW and 900 W respectively. The friction and windage losses are 1050 W and the core losses are 1200 W. The air-gap power of the motor will be (A) 23.06 kW (B) 24.11 kW (C) 25.01 kW

EE SP 3.62

TWO MARKS

(D) 26.21 kW

The core of a two-winding transformer is subjected to a magnetic flux variation as indicated in the figure.

The induced emf (ers) in the secondary winding as a function of time will be of the form


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Electric Machines

Page 141

Common Data for Questions 63 and 64 A 3-phase, 440 V, 50 Hz, 4-pole slip ring induction motor is feed from the rotor side through an auto-transformer and the stator is connected to a variable resistance as shown in the figure.

nodia

The motor is coupled to a 220 V, separately excited d.c generator feeding power to fixed resistance of 10 W. Two-wattmeter method is used to measure the input power to induction motor. The variable resistance is adjusted such the motor runs at 1410 rpm and the following readings were recorded W1 = 1800 W, W2 =- 200 W.

EE SP 3.63

The speed of rotation of stator magnetic field with respect to rotor structure will be (A) 90 rpm in the direction of rotation (B) 90 rpm in the opposite direction of rotation (C) 1500 rpm in the direction of rotation (D) 1500 rpm in the opposite direction of rotation

EE SP 3.64

Neglecting all losses of both the machines, the dc generator power output and the current through resistance (Rex) will respectively be (A) 96 W, 3.10 A (B) 120 W, 3.46 A (C) 1504 W, 12.26 A (D) 1880 W, 13.71 A

EE SP 3.65

A 400 V, 50 Hz, 4-pole, 1400 rpm, star connected squirrel cage induction motor has the following parameters referred to the stator: R'r = 1.0 W, Xs = X'r = 1.5 W Neglect stator resistance and core and rotational losses of the motor. The motor is controlled from a 3-phase voltage source inverter with constant V/f control. The stator line-to-line voltage(rms) and frequency to obtain the maximum torque at starting will be : (A) 20.6 V, 2.7 Hz (B) 133.3 V, 16.7 Hz (C) 266.6 V, 33.3 Hz (D) 323.3 V, 40.3 Hz

Statement for Linked Answer Question 66 and 67 A 240 V, dc shunt motor draws 15 A while supplying the rated load at a speed of 80 rad/s. The armature resistance is 0.5 W and the field winding resistance is


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Electric Machines

Chapter 3

80 W. EE SP 3.66

The net voltage across the armature resistance at the time of plugging will be (A) 6 V (B) 234 V (C) 240 V (D) 474 V

EE SP 3.67

The external resistance to be added in the armature circuit to limit the armature current to 125% of its rated value is (A) 31.1 W (B) 31.9 W (C) 15.1 W (D) 15.9 W


nodia

A synchronous motor is connected to an infinite bus at 1.0 pu voltage and draws 0.6 pu current at unity power factor. Its synchronous reactance is 1.0 pu resistance is negligible.

EE SP 3.68

EE SP 3.69

The excitation voltage (E ) and load angle (d) will respectively be (A) 0.8 pu and 36.86c lag (B) 0.8 pu and 36.86c lead (C) 1.17 pu and 30.96c lead (D) 1.17 pu and 30.96c lag Keeping the excitation voltage same, the load on the motor is increased such that the motor current increases by 20%. The operating power factor will become (A) 0.995 lagging (B) 0.995 leading (C) 0.791 lagging (D) 0.848 leading YEAR 2007

ONE MARK

EE SP 3.70

In a transformer, zero voltage regulation at full load is (A) not possible (B) possible at unity power factor load (C) possible at leading power factor load (D) possible at lagging power factor load

EE SP 3.71

The dc motor, which can provide zero speed regulation at full load without any controller is (A) series (B) shunt (C) cumulative compound (D) differential compound

EE SP 3.72

The electromagnetic torque Te of a drive and its connected load torque TL are as shown below. Out of the operating points A, B, C and D, the stable ones are


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Electric Machines

Page 143

(A) A, C, D

(B) B, C

(C) A, D

(D) B, C, D

nodia YEAR 2007

EE SP 3.73

TWO MARKS

A three-phase synchronous motor connected to ac mains is running at full load and unity power factor. If its shaft load is reduced by half, with field current held constant, its new power factor will be (A) unity (B) leading (C) lagging (D) dependent on machine parameters

EE SP 3.74

A 100 kVA, 415 V(line), star-connected synchronous machine generates rated open circuit voltage of 415 V at a field current of 15 A. The short circuit armature current at a field current of 10 A is equal to the rated armature current. The per unit saturated synchronous reactance is (A) 1.731 (B) 1.5 (C) 0.666 (D) 0.577

EE SP 3.75

A single-phase, 50 kVA, 250 V/500 V two winding transformer has an efficiency of 95% at full load, unity power factor. If it is re-configured as a 500 V/750 V auto-transformer, its efficiency at its new rated load at unity power factor will be (A) 95.752% (B) 97.851% (C) 98.276% (D) 99.241%

EE SP 3.76

A three-phase, three-stack, variable reluctance step motor has 20 poles on each rotor and stator stack. The step angle of this step motor is (A) 3c (B) 6c (C) 9c (D) 18c

EE SP 3.77

A three-phase squirrel cage induction motor has a starting torque of 150% and a maximum torque of 300% with respect to rated torque at rated voltage and


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Electric Machines

Chapter 3

rated frequency. Neglect the stator resistance and rotational losses. The value of slip for maximum torque is (A) 13.48% (B) 16.42% (C) 18.92% (D) 26.79%

Common Data for Question 78 to 80 A three phase squirrel cage induction motor has a starting current of seven times the full load current and full load slip of 5% EE SP 3.78

EE SP 3.79

EE SP 3.80

If an auto transformer is used for reduced voltage starting to provide 1.5 per unit starting torque, the auto transformer ratio(%) should be (A) 57.77 % (B) 72.56 % (C) 78.25 % (D) 81.33 %

nodia

If a star-delta starter is used to start this induction motor, the per unit starting torque will be (A) 0.607 (B) 0.816 (C) 1.225 (D) 1.616 If a starting torque of 0.5 per unit is required then the per unit starting current should be (A) 4.65 (B) 3.75 (C) 3.16

(D) 2.13

YEAR 2006

ONE MARK

EE SP 3.81

In transformers, which of the following statements is valid ? (A) In an open circuit test, copper losses are obtained while in short circuit test, core losses are obtained (B) In an open circuit test, current is drawn at high power factor (C) In a short circuit test, current is drawn at zero power factor (D) In an open circuit test, current is drawn at low power factor

EE SP 3.82

For a single phase capacitor start induction motor which of the following statements is valid ? (A) The capacitor is used for power factor improvement (B) The direction of rotation can be changed by reversing the main winding terminals (C) The direction of rotation cannot be changed (D) The direction of rotation can be changed by interchanging the supply terminals

EE SP 3.83

In a DC machine, which of the following statements is true ? (A) Compensating winding is used for neutralizing armature reaction while interpole winding is used for producing residual flux


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Electric Machines

Page 145

(B) Compensating winding is used for neutralizing armature reaction while interpole winding is used for improving commutation (C) Compensating winding is used for improving commutation while interpole winding is used for neutralizing armature reaction (D) Compensation winding is used for improving commutation while interpole winding is used for producing residual flux YEAR 2006 EE SP 3.84

TWO MARKS

A 220 V DC machine supplies 20 A at 200 V as a generator. The armature resistance is 0.2 ohm. If the machine is now operated as a motor at same terminal voltage and current but with the flux increased by 10%, the ratio of motor speed to generator speed is (A) 0.87 (B) 0.95

nodia (C) 0.96

EE SP 3.85

(D) 1.06

A synchronous generator is feeding a zero power factor (lagging) load at rated current. The armature reaction is (A) magnetizing (B) demagnetizing (C) cross-magnetizing (D) ineffective

EE SP 3.86

Two transformers are to be operated in parallel such that they share load in proportion to their kVA ratings. The rating of the first transformer is 500 kVA ratings. The rating of the first transformer is 500 kVA and its pu leakage impedance is 0.05 pu. If the rating of second transformer is 250 kVA, its pu leakage impedance is (A) 0.20 (B) 0.10 (C) 0.05 (D) 0.025

EE SP 3.87

The speed of a 4-pole induction motor is controlled by varying the supply frequency while maintaining the ratio of supply voltage to supply frequency ( V/f ) constant. At rated frequency of 50 Hz and rated voltage of 400 V its speed is 1440 rpm. Find the speed at 30 Hz, if the load torque is constant (A) 882 rpm (B) 864 rpm (C) 840 rpm (D) 828 rpm

EE SP 3.88

A 3-phase, 4-pole, 400 V 50 Hz , star connected induction motor has following circuit parameters r1 = 1.0 W, r'2 = 0.5 W, X1 = X'2 = 1.2 W, Xm = 35 W The starting torque when the motor is started direct-on-line is (use approximate equivalent circuit model) (A) 63.6 Nm (B) 74.3 Nm (C) 190.8 Nm (D) 222.9 Nm


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Electric Machines

Chapter 3

EE SP 3.89

A 3-phase, 10 kW, 400 V, 4-pole, 50Hz, star connected induction motor draws 20 A on full load. Its no load and blocked rotor test data are given below. No Load Test : 400 V 6A 1002 W Blocked Rotor Test : 90 V 15 A 762 W Neglecting copper loss in no load test and core loss in blocked rotor test, estimate motor’s full load efficiency (A) 76% (B) 81% (C) 82.4% (D) 85%

EE SP 3.90

A 3-phase, 400 V, 5 kW, star connected synchronous motor having an internal reactance of 10 W is operating at 50% load, unity p.f. Now, the excitation is increased by 1%. What will be the new load in percent, if the power factor is to be kept same ? Neglect all losses and consider linear magnetic circuit. (A) 67.9% (B) 56.9%

nodia (C) 51%

(D) 50%

Common Data For Q. 91 and 92

A 4-pole, 50 Hz, synchronous generator has 48 slots in which a double layer winding is housed. Each coil has 10 turns and is short pitched by an angle to 36c electrical. The fundamental flux per pole is 0.025 Wb

EE SP 3.91

The line-to-line induced emf(in volts), for a three phase star connection is approximately (A) 808 (B) 888 (C) 1400 (D) 1538

EE SP 3.92

The line-to-line induced emf(in volts), for a three phase connection is approximately (A) 1143 (B) 1332 (C) 1617 (D) 1791

EE SP 3.93

The fifth harmonic component of phase emf(in volts), for a three phase star connection is (A) 0 (B) 269 (C) 281 (D) 808

Statement for Linked Answer Questions 94 and 95 A 300 kVA transformer has 95% efficiency at full load 0.8 p.f. lagging and 96% efficiency at half load, unity p.f. EE SP 3.94

The iron loss (Pi) and copper loss (Pc) in kW, under full load operation are (B) Pc = 6.59, Pi = 9.21 (A) Pc = 4.12, Pi = 8.51 (C) Pc = 8.51, Pi = 4.12 (D) Pc = 12.72, Pi = 3.07

EE SP 3.95

What is the maximum efficiency (in %) at unity p.f. load ?


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Electric Machines

Page 147

(A) 95.1

(B) 96.2

(C) 96.4

(D) 98.1


EE SP 3.97

The equivalent circuit of a transformer has leakage reactances X1, X'2 and magnetizing reactance XM . Their magnitudes satisfy (A) X1 >> X'2 >> XM

(B) X1 << X'2 << XM

(C) X1 . X'2 >> XM

(D) X1 . X'2 << XM

Which three-phase connection can be used in a transformer to introduce a phase difference of 30c between its output and corresponding input line voltages (A) Star-Star (B) Star-Delta

nodia (C) Delta-Delta

EE SP 3.98

(D) Delta-Zigzag

On the torque/speed curve of the induction motor shown in the figure four points of operation are marked as W, X, Y and Z. Which one of them represents the operation at a slip greater than 1 ?

(A) W (C) Y EE SP 3.99

ONE MARK

(B) X (D) Z

For an induction motor, operation at a slip s , the ration of gross power output to air gap power is equal to (A) (1 - s) 2 (B) (1 - s) (C)

(1 - s)

(D) (1 -

s)

YEAR 2005 EE SP 3.100

TWO MARKS

Two magnetic poles revolve around a stationary armature carrying two coil (c1 - c1l , c2 - c2l ) as shown in the figure. Consider the instant when the poles are in a position as shown. Identify the correct statement regarding the polarity of the induced emf at this instant in coil sides c1 and c2 .


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Electric Machines

Chapter 3

(A) 9 in c1 , no emf in c2 (B) 7 in c1 , no emf in c2 (C) 9 in c2 , no emf in c1 (D) 7 in c2 , no emf in c1 EE SP 3.101

nodia

A 50 kW dc shunt is loaded to draw rated armature current at any given speed. When driven (i) at half the rated speed by armature voltage control and (ii) at 1.5 times the rated speed by field control, the respective output powers delivered by the motor are approximately. (A) 25 kW in (i) and 75 kW in (ii) (B) 25 kW in (i) and 50 kW in (ii) (C) 50 kW in (i) and 75 kW in (ii) (D) 50 kW in (i) and 50 kW in (ii)

EE SP 3.102

In relation to the synchronous machines, which on of the following statements is false ? (A) In salient pole machines, the direct-axis synchronous reactance is greater than the quadrature-axis synchronous reactance. (B) The damper bars help the synchronous motor self start. (C) Short circuit ratio is the ratio of the field current required to produces the rated voltage on open circuit to the rated armature current. (D) The V-cure of a synchronous motor represents the variation in the armature current with field excitation, at a given output power.

EE SP 3.103

In relation to DC machines, match the following and choose the correct combination List-I

List-II

Performance Variables

Proportional to

P. Armature emf (E )

1. Flux(f), speed (w) and armature current (Ia)

Q. Developed torque (T )

2. f and w only

R Developed power (P )

3. f and Ia only 4. Ia and w only 5. Ia only

Codes: (A) (B)

P 3 2

Q 3 5

R 1 4


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Electric Machines

(C) (D) EE SP 3.104

3 2

5 3

Page 149

4 1

Under no load condition, if the applied voltage to an induction motor is reduced from the rated voltage to half the rated value, (A) the speed decreases and the stator current increases (B) both the speed and the stator current decreases (C) the speed and the stator current remain practically constant (D) there is negligible change in the speed but the stator current decreases

EE SP 3.105

A three-phase cage induction motor is started by direct-on-line (DOL) switching at the rated voltage. If the starting current drawn is 6 times the full load current, and the full load slip is 4%, then ratio of the starting developed torque to the full load torque is approximately equal to (A) 0.24 (B) 1.44

nodia (C) 2.40

EE SP 3.106

EE SP 3.107

(D) 6.00

In a single phase induction motor driving a fan load, the reason for having a high resistance rotor is to achieve (A) low starting torque (B) quick acceleration (C) high efficiency (D) reduced size

Determine the correctness or otherwise of the following assertion[A] and the reason[R] Assertion [A] : Under V/f control of induction motor, the maximum value of the developed torque remains constant over a wide range of speed in the subsynchronous region. Reason [R] : The magnetic flux is maintained almost constant at the rated value by keeping the ration V/f constant over the considered speed range. (A) Both [A] and [R] are true and [R] is the correct reason for [A] (B) Both [A] and [R] are true and but [R] is not the correct reason for [A] (C) Both [A] and [R] are false (D) [A] is true but [R] is false

Statement for Linked Answer Questions 108 and 109 A 1000 kVA, 6.6 kV, 3-phase star connected cylindrical pole synchronous generator has a synchronous reactance of 20 W. Neglect the armature resistance and consider operation at full load and unity power factor. EE SP 3.108

The induced emf(line-to-line) is close to (A) 5.5 kV (B) 7.2 kV (C) 9.6 kV (D) 12.5 kV

EE SP 3.109

The power(or torque) angle is close to (A) 13.9c (C) 24.6c

(B) 18.3c (D) 33.0c


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Electric Machines

Chapter 3

YEAR 2004 EE SP 3.110

ONE MARK

A 500 kVA, 3-phase transformer has iron losses of 300 W and full load copper losses of 600 W. The percentage load at which the transformer is expected to have maximum efficiency is (A) 50.0% (B) 70.7% (C) 141.4%

EE SP 3.111

EE SP 3.112

(D) 200.0%

For a given stepper motor, the following torque has the highest numerical value (A) Detent torque (B) Pull-in torque (C) Pull-out torque (D) Holding torque

nodia The following motor definitely has a permanent magnet rotor (A) DC commutator motor (B) Brushless dc motor (C) Stepper motor (D) Reluctance motor

EE SP 3.113

The type of single-phase induction motor having the highest power factor at full load is (A) shaded pole type (B) split-phase type (C) capacitor-start type

(D) capacitor-run type

EE SP 3.114

The direction of rotation of a 3-phase induction motor is clockwise when it is supplied with 3-phase sinusoidal voltage having phase sequence A-B-C. For counter clockwise rotation of the motor, the phase sequence of the power supply should be (A) B-C-A (B) C-A-B (C) A-C-B (D) B-C-A or C-A-B

EE SP 3.115

For a linear electromagnetic circuit, the following statement is true (A) Field energy is equal to the co-energy (B) Field energy is greater than the co-energy (C) Field energy is lesser than the co-energy (D) Co-energy is zero YEAR 2004

EE SP 3.116

TWO MARKS

The synchronous speed for the seventh space harmonic mmf wave of a 3-phase, 8-pole, 50 Hz induction machine is (A) 107.14 rpm in forward direction (B) 107.14 rpm in reverse direction (C) 5250 rpm in forward direction


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Electric Machines

Page 151

(D) 5250 rpm in reverse direction EE SP 3.117

A rotating electrical machine its self-inductances of both the stator and the rotor windings, independent of the rotor position will be definitely not develop (A) starting torque (B) synchronizing torque (C) hysteresis torque (D) reluctance torque

EE SP 3.118

The armature resistance of a permanent magnet dc motor is 0.8 W. At no load, the motor draws 1.5 A from a supply voltage of 25 V and runs at 1500 rpm. The efficiency of the motor while it is operating on load at 1500 rpm drawing a current of 3.5 A from the same source will be (A) 48.0% (B) 57.1%

nodia (C) 59.2%

EE SP 3.119

(D) 88.8%

A 50 kVA, 3300/230 V single-phase transformer is connected as an autotransformer shown in figure. The nominal rating of the autotransformer will be

(A) 50.0 kVA (C) 717.4 kVA

(B) 53.5 kVA (D) 767.4 kVA

EE SP 3.120

The resistance and reactance of a 100 kVA, 11000/400 V, 3- Y distribution transformer are 0.02 and 0.07 pu respectively. The phase impedance of the transformer referred to the primary is (A) (0.02 + j0.07) W (B) (0.55 + j1.925) W (C) (15.125 + j52.94) W (D) (72.6 + j254.1) W

EE SP 3.121

A single-phase, 230 V, 50 Hz 4-pole, capacitor-start induction motor had the following stand-still impedances Auxiliary winding Za = 8.0 + j6.0 W Main winding Zm = 6.0 + j4.0 W , The value of the starting capacitor required to produce 90c phase difference between the currents in the main and auxiliary windings will be (A) 176.84 mF (B) 187.24 mF (C) 265.26 mF (D) 280.86 mF

EE SP 3.122

Two 3-phase, Y-connected alternators are to be paralleled to a set of common busbars. The armature has a per phase synchronous reactance of 1.7 W and negligible armature resistance. The line voltage of the first machine is adjusted to 3300 V and that of the second machine is adjusted to 3200 V. The machine voltages are in phase at the instant they are paralleled. Under this condition, the


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Electric Machines

Chapter 3

synchronizing current per phase will be (A) 16.98 A (B) 29.41 A (C) 33.96 A (D) 58.82 A EE SP 3.123

A 400 V, 15 kW, 4-pole, 50Hz, Y-connected induction motor has full load slip of 4%. The output torque of the machine at full load is (A) 1.66 Nm (B) 95.50 Nm (C) 99.47 Nm (D) 624.73 Nm

EE SP 3.124

For a 1.8c, 2-phase bipolar stepper motor, the stepping 100 steps/second. The rotational speed of the motor in rpm is (A) 15 (B) 30 (C) 60 (D) 90

EE SP 3.125

EE SP 3.127

is

nodia

A 8-pole, DC generator has a simplex wave-wound armature containing 32 coils of 6 turns each. Its flux per pole is 0.06 Wb. The machine is running at 250 rpm. The induced armature voltage is (A) 96 V (B) 192 V (C) 384 V

EE SP 3.126

rate

(D) 768 V

A 400 V, 50 kVA, 0.8 p.f. leading 3-connected, 50 Hz synchronous machine has a synchronous reactance of 2 W and negligible armature resistance. The friction and windage losses are 2 kW and the core loss is 0.8 kW. The shaft is supplying 9 kW load at a power factor of 0.8 leading. The line current drawn is (A) 12.29 A (B) 16.24 A (C) 21.29 A (D) 36.88 A A 500 MW, 3-phase, Y-connected synchronous generator has a rated voltage of 21.5 kV at 0.85 p.f. The line current when operating at full load rated conditions will be (A) 13.43 kA (B) 15.79 kA (C) 23.25 kA (D) 27.36 kA YEAR 2003

ONE MARK

EE SP 3.128

A simple phase transformer has a maximum efficiency of 90% at full load and unity power factor. Efficiency at half load at the same power factor is (A) 86.7% (B) 88.26% (C) 88.9% (D) 87.8%

EE SP 3.129

Group-I lists different applications and Group-II lists the motors for these applications. Match the application with the most suitable motor and choose the right combination among the choices given thereafter Group-I

Group-II


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Electric Machines

Page 153

P.

Food mixer

1. Permanent magnet dc motor

Q

Cassette tape recorder

2. Single-phase induction motor

R.

Domestic water pump

3. Universal motor

S.

Escalator

4. Three-phase induction motor 5. DC series motor 6. Stepper motor

Codes: (A) (B) (C) (D)

EE SP 3.130

P 3 1 3 3

Q 6 3 1 2

R 4 2 2 1

S 5 4 4 4

nodia

A stand alone engine driven synchronous generator is feeding a partly inductive load. A capacitor is now connected across the load to completely nullify the inductive current. For this operating condition. (A) the field current and fuel input have to be reduced (B) the field current and fuel input have to be increased

(C) the field current has to be increased and fuel input left unaltered (D) the field current has to be reduced and fuel input left unaltered

EE SP 3.131

Curves X and Y in figure denote open circuit and full-load zero power factor(zpf) characteristics of a synchronous generator. Q is a point on the zpf characteristics at 1.0 p.u. voltage. The vertical distance PQ in figure gives the voltage drop across

(A) Synchronous reactance (C) Potier reactance EE SP 3.132

(B) Magnetizing reactance (D) Leakage reactance

No-load test on a 3-phase induction motor was conducted at different supply voltage and a plot of input power versus voltage was drawn. This curve was extrapolated to intersect the y-axis. The intersection point yields (A) Core loss (B) Stator copper loss (C) Stray load loss (D) Friction and windage loss


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Electric Machines

YEAR 2003 EE SP 3.133

Chapter 3

TWO MARKS

Figure shows an ideal single-phase transformer. The primary and secondary coils are wound on the core as shown. Turns ratio N1 /N2 = 2 .The correct phasors of voltages E1, E2 , currents I1, I2 and core flux F are as shown in

nodia EE SP 3.134

To conduct load test on a dc shunt motor, it is coupled to a generator which is identical to the motor. The field of the generator is also connected to the same supply source as the motor. The armature of generator is connected to a load resistance. The armature resistance is 0.02 p.u. Armature reaction and mechanical losses can be neglected. With rated voltage across the motor, the load resistance across the generator is adjusted to obtain rated armature current in both motor and generator. The p.u value of this load resistance is (A) 1.0 (B) 0.98 (C) 0.96 (D) 0.94

EE SP 3.135

Figure shows a 3- Y connected, 3-phase distribution transformer used to step down the voltage from 11000 V to 415 V line-to-line. It has two switches S1 and S 2 . Under normal conditions S1 is closed and S 2 is open. Under certain special conditions S1 is open and S 2 is closed. In such a case the magnitude of the voltage across the LV terminals a and c is


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EE SP 3.136

Electric Machines

Page 155

(A) 240 V

(B) 480 V

(C) 415 V

(D) 0 V

Figure shows an ideal three-winding transformer. The three windings 1, 2, 3 of the transformer are wound on the same core as shown. The turns ratio N1: N2: N3 is 4: 2: 1. A resistor of 10 W is connected across winding-2. A capacitor of reactance 2.5 W is connected across winding-3. Winding-1 is connected across a 400 V, ac supply. If the supply voltage phasor V1 = 400+0% , the supply current phasor I1 is given by

nodia (A) (- 10 + j10) A (C) (10 + j10) A EE SP 3.137

EE SP 3.138

(B) (- 10 - j10) A (D) (10 - j10) A

Following are some of the properties of rotating electrical machines P. Stator winding current is dc, rotor winding current is ac. Q. Stator winding current is ac, rotor winding current is dc. R. Stator winding current is ac, rotor winding current is ac. S. Stator has salient poles and rotor has commutator. T. Rotor has salient poles and sliprings and stator is cylindrical. U. Both stator and rotor have poly-phase windings. DC machines, Synchronous machines and Induction machines exhibit some of the above properties as given in the following table. Indicate the correct combination from this table DC Machine

Synchronous Machines

Induction Machines

(A) P,S (B) Q,U (C) P,S (D) R,S

Q,T P,T R,U Q,U

R,U R,S Q,T P,T

When stator and rotor windings of a 2-pole rotating electrical machine are excited, each would produce a sinusoidal mmf distribution in the airgap with peal values Fs and Fr respectively. The rotor mmf lags stator mmf by a space angle d at any instant as shown in figure. Thus, half of stator and rotor surfaces will form one pole with the other half forming the second pole. Further, the direction of torque


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Electric Machines

Chapter 3

acting on the rotor can be clockwise or counter-clockwise.

The following table gives four set of statement as regards poles and torque. Select the correct set corresponding to the mmf axes as shown in figure.

EE SP 3.139

nodia Stator Surface ABC forms

Stator Surface CDA forms

Rotor Surface abc forms

Rotor surface cda forms

Torque is

(A) North Pole (B) South Pole

South Pole North Pole

North Pole North Pole

South Pole South Pole

(C) North Pole

South Pole

South Pole

North Pole

(D) South Pole

North Pole

South Pole

North Pole

Clockwise Counter Clockwise Counter Clockwise Clockwise

A 4-pole, 3-phase, double-layer winding is housed in a 36-slot stator for an ac machine with 60c phase spread. Coil span is 7 short pitches. Number of slots in which top and bottom layers belong to different phases is (A) 24 (B) 18 (C) 12 (D) 0

EE SP 3.140

A 3-phase induction motor is driving a constant torque load at rated voltage and frequency. If both voltage and frequency are halved, following statements relate to the new condition if stator resistance, leakage reactance and core loss are ignored 1. The difference between synchronous speed and actual speed remains same 2. The airgap flux remains same 3. The stator current remains same 4. The p.u. slip remains same Among the above, current statements are (A) All (B) 1, 2 and 3 (C) 2, 3 and 4 (D) 1 and 4

EE SP 3.141

A single-phase induction motor with only the main winding excited would exhibit the following response at synchronous speed


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Electric Machines

Page 157

(A) Rotor current is zero (B) Rotor current is non-zero and is at slip frequency (C) Forward and backward rotaling fields are equal (D) Forward rotating field is more than the backward rotating field EE SP 3.142

A dc series motor driving and electric train faces a constant power load. It is running at rated speed and rated voltage. If the speed has to be brought down to 0.25 p.u. the supply voltage has to be approximately brought down to (A) 0.75 p.u (B) 0.5 p.u (C) 0.25 p.u

(D) 0.125 p.u

nodia ***********


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Electric Machines

Chapter 3

SOLUTION SOL 3.1

Correct answer is 3.33 Hz. For the given induction motor, we have P = 8 , f = 50 Hz , Nr = 700 rpm So, we obtain Ns =

120f = 120 # 50 = 750 rpm 8 P

Therefore, the slip s = Ns - Nr = 750 - 700 = 1 15 750 Ns Thus, the frequency of rotor current of the motor is obtained as

nodia Frequency of rotor current = s # f = 1 # 50 = 3.33 Hz 15

SOL 3.2

Correct option is (D). EMF equation of Transformer is given by E = 2 pffN where f is the flux density, given as

f = B#A Now, the given radius is reduced by half, i.e. r " r/2 So, the area will be reduced by 1/4, i.e.

A " A/4 To maintain no load current constant [in turn emf (E ) to be constant], number of turn should be increased by 4 times. SOL 3.3

Correct answer is 15.06 W . For the open circuit test, we have line-to-line voltage at I f = 2.3 A as Voc = 400 V So, the open circuit phase voltage is obtained as Voc = 400 V 3 I = 2.3 A Again, for short circuit test, we have phase rms current at I f = 1.5 A as Isc = 10 A I = 1.5 A L-L

Ph

g

Ph

g

Since Isc and I f are linearly related, so we get the short circuit phase current at I f = 2.3 A as Iscl = 10 # 2.3 = 46 A 1.5 3 I = 2.3 A Thus, per phase synchronous impedance is V Z ph = oc Iscl I = 2.3 A Ph

g

Ph

Ph

=

g

400/ 3 = 400 # 46 46/3

3 = 15.06 W


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Electric Machines

Page 159

Correct option is (A). For V = 230 V and f = 50 Hz , we have Core loss = 1050 W or Iron loss + Hysteresis loss = 1050 W Since, iron loss and hysteresis loss are given by

...(i)

Iron loss = Pi = ki V 2 Hysteresis loss = PH = KH V 1.6 f - 0.6 Substituting it in equation (i), we get ^230h1.6 Ki ^230h2 + KH = 1050 ^50h0.6 or Ki ^52900h + KH ^574.62h = 1050 Again, for V = 138 V and f = 30 Hz , we have and

nodia or or

...(ii)

Core loss = 500 W Iron loss + Hysteresis loss = 500 W ^138h1.6 = 500 KP ^138h2 + KH ^30h0.6

or KP ^19044h + KH ^344.77h = 500 Solving equations (i) and (ii), we get

and For 230 V, 50 Hz; we have

...(ii)

KP = 0.01024 KH = 0.855

Hysteresis loss = KH V 1.6 f - 0.6 = 0.855 # ^230h1.6 # ^50h-0.6 = 508.53 W Hence, the eddy loss is Eddy loss = KIV = 0.01024 # 52900 2

= 541.69 W SOL 3.5

Correct answer is 1241. rpm. For the given data, we draw the circuit as

So, the current through field resistance is I f = V = 230 = 1 A 230 Rf Hence, we have the armature current, Ia = I - I f = 5 - 1 = 4 A


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Electric Machines

Chapter 3

From the given circuit, we have or

V = E f 1 + Ia # Rs 230 = E f 1 + 4 # 0.4

or At full load,

E f 1 = 228.4 V I = 70 A

So, Ia = 70 - 1 = 69 A and E f 2 = 230 - 69 # 0.4 = 202.4 V as emf \ flux # speed For DC shunt motor flux ^fh is constant. So, Ef \ w Ef 1 or = w1 w2 Ef 2 228.4 = 1400 or w2 202.4 Thus, w 2 = 202.4 # 1400 = 1241.1 rpm 228.4

SOL 3.6

nodia Correct option is (C). For the induction motor, we have

s f R2 R + ^s f X2h2 or Tmax ^maxm torqueh \ 1 2X2 Tf 2s R X So, we get = 2 f 2 2 2 Tmax R 2 + ^s f X2h Tf 2 or = Tmax s max + s f sf s max Now, the slip at full load is 3% , i.e. Tf (Full load torque) \

2 2

...(i)

s f = 0.003 and s max = R2 = 0.1 = 0.108 X2 0.92 Substituting these values in equation (i), we get Tf 2 = 0.108 + 0.03 Tmax 0.03 0.108 Tmax = 3.623 + 0.276 or 2 Tf Tmax = 1.948 or Tf i.e. the ratio of maximum torque to full load torque is 1.948. SOL 3.7



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Electric Machines

Page 161

Referring to secondary side

VTh = b N2 l Vs N1 VTh = 2 sin ^wt h = 2 sin ^wt h 1 2 RTh = b 2 l RP 1 2 = b 2 l RP = 4 # 1 = 4 W 1

SOL 3.8

nodia

Correct answer is 550. From the given information for the two winding transformer, we have V1 = 1000 V , V2 = 100 V , S 0 = 50 kVA So, I1 = 50 A , I2 = 500 A For the autotransformer as shown in figure

Vi = V1 = 1000 V Vo = V1 + V2 = 1000 + 100 = 1100 V Rating of auto transformer S 0a = V0 # I2a = 1100 # 500 = 550 kVA SOL 3.9

Î2a = I2h



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Electric Machines

Chapter 3

As the load decreases, the speed increases. Relative speed between stator rmf and rotor speed reduced which decrease the slip consequently emf induced in the rotor and current decreases. Hence frequency decreases.

SOL 3.10

So

f 1, f 2 < 50 Hz

but

f2 > f1

Correct answer is 7. Let complex power (kVA) is given as S = P + jQ P " Real power ^W h Q " Reactive Power ^VR h P = S cos q Q = S sin q or Q = P sin q = P tan q cos q Initially when the real power is 12 MW and power factor is 0.6, the reactive power is Real power Reactive power

nodia Q1 = P1 tan q 1

q = cos-1 ^0.6h = 53.13c = 12 tan ^53.13ch = 12 # 4 = 16 MVR 3 Now, when capacitor is added, improved power factor angle Q2 = cos-1 ^0.80h = 36.86c Total reactive power after adding capacitor

Q2 = P1 tan q 2 Q2 = 12 # tan ^36.86ch = 12 # 3 = 9 MAR 4 From the power triangle (before and after capacitor is added)

(P1 is same)

Reactive power delivered by capacitor QC = Q1 - Q2 = 16 - 9 = 7 MVR


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Electric Machines

Page 163

Correct option is (B). V = Constant f At low frequency the stator plan decrease from its rated value as at low frequency will increase the flux density which results in saturation of stator are therefore stator flux decreases.

SOL 3.12

Correct answer is 1.27 . When the machine is running as a generator Field current Isn = 250 = 2 A 120 Armature current

IL = 50 A Ia = IL + Isn = 50 + 2 = 52 A

Load current Armature current

IlL = 50 A Ila = I lL - I lsn = 50 - 2 = 48 A

Load current

nodia Induced Emf Eg = 250 + 52 ^0.6h = 281.2 V When the machine is running as a motor Field current Ilsn = 250 = 2 A 125

Induced Emf Em = 250 - 48 ^0.6h = 221.2 V For DC machine, f is constant, so Ng = Nm Em Eg Ng E Speed Ratio = g = 281.2 = 1.27 Nm Em 221.2

SOL 3.13

Correct option is (A). Given that Rotor winding is open circuit and rotor is rotating anti-clockwise, therefore Slip s = Ns + N r Ns or f 1 = fs + f r = f + fr So frequency of voltage across slip rings is ^ f + fr h Hz . At the same time, frequency of voltage across commutator brushes is same as rotor frequency f2 = f

SOL 3.14

Correct option is (D). For single phase winding no. of coils in the phase n = 180 = 9 20 # 1 The slop span in electric degree 180 = 180 = 20c no. of slots per pole 180/20 sin ^nr/2h sin ^9 # 20/2h = = 0.6398 kd1 = n sin ^r/2h 9 # sin ^20/2h N1 = no. of slots # no. of conductors = 180 # 6 = 540 2 2 r =

Distribution factor No. of turns,


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Electric Machines

Chapter 3

Now for three phase winding Slot span No. of coils in a phase group

r = 20c

180 = 3 20 # 3 sin ^nr/2h sin ^3 # 20/2h = = 0.9597 kd 2 = n sin ^r/2h 3 # sin ^20/2h 1 N2 = no. of slots # no. of conductor # 2 no. of phase = 180 # 6 # 1 2 3 n =

Distribution factor no. of turns,

= 180 The rms value of generator voltage

nodia V1 = 4.44 kd 1 ffN1 V2 = 4.44kd2 ffN2 V1 = kd 1 # N1 = 0.6398 # 540 0.9597 # 180 V2 kd 2 # N2

For single, phase, For three phase,

-2

SOL 3.15

Correct option is (A). Hysteris loss in transformer

n Ph = kn fB max Where, the kn is the proportionality constant dependent on the characteristics and volume of material, n is known as steinmetz exponents. B max \ V f Change in voltage and frequency is same, so B max is constant. So, Ph \ f Change in voltage and frequency is same, so B max is constant. So, Ph \ f Therefore percentage change in hysteris loss will be same as that of change in frequency

TPh = Tf = 10% Eddy current loss is given by

Pe = ke ^B max f d h2

Where d = B max = f = ke = Again, So,

lamination thickness maximum flux density frequency Proportionality constant (Constant) B max \ V f Pe \ f 2 2 Pe1 = f 1 Pe2 f 22 f1 f = = 1 1.1 f2 f + 0.1f


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Electric Machines

Page 165

Pe1 = 1 2 Pe2 b 1.1 l

So

Pe2 = ^1.1h2 Pe1 = 1.21Pe1 Now, percentage change in eddy current loss TPe = Pe2 - Pe1 # 100 Pe1 = 1.21 - 1 # 100 1

or

= 21% SOL 3.16

SOL 3.17

Correct option is (D). After sudden change of the load in three phase synchronous machine, the rotor has to search or hunt for its new equilibrium position. It causes the hunting, which can be damped by rotor copper losses.

nodia

Correct answer is (C). If the switch operates correctly, then the torque-speed characteristic is

But, the switch does not operate due to malfunctioning. So, there will be no discontinuity in the characteristic curve. Thus, we have the modified characteristic curve of the motor as

SOL 3.18

Correct answer is 12.55 N-m. A dc motor circuit is shown below.


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Electric Machines

Chapter 3

Given No load speed, Armature current,

N = 1400 rpm Ia = 8 A

V = 230 V So, we obtain the torque of the motor as t = V Ia = 2pV Ia w 60 # N Excitation voltage,

= 2p230 # 8 60 # 1400 = 9.55 # 230 # 8 1400 = 12.55 N-m

SOL 3.19

SOL 3.20

nodia

Correct option is (B). Since, increase in torque accelerates the motor, so the slope of torque speed characteristic should be negative to restore the speed to original value, which in turn makes the system stable. Thus, for stable operating point, the slope of torque-speed characteristic should be negative. Hence, the points P and S are stable operating points. Correct option is (B). Since, we know that

where

PC = PH f + Pe f 2 PC = total core losses of the transformer f = frequency PH = Hysteresis losses constant

...(i)

Pe = eddy current losses constant Now, we have the plot of core loss/frequency versus frequency as

So, we rewrite equation (i) as PC = P + P f ...(ii) H e f i.e. the plot between PC /f and f is a straight line. From the graph, we have the values of PH and Pe as


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Electric Machines

PH = 10 ;

Page 167

Pe = 1 10

From equation (i), we have the hysteresis and eddy current losses as Pe =

PH f = 10 # 25 = 250 W Pe f 2 = 1 # ^25h2 = 62.5 W 10

Thus, we get

SOL 3.21

PH f + Pe f 2 S S Hysteresis loss Eddy current loss

Correct answer is 33.47°. For the non-salient synchronous generator, we have Synchronous reactance, Supplied power, Power, factor,

XS = 0.8 p.u. P = 1 p.u.

nodia

cos f = 1 Terminal voltage, Vt = 1.1 p.u. Since, power factor is unity, so phase angle of armature current is also zero. So, we get 1 = = 1 Ia = P Vt cos f (1.1) (1) 1.1 or Ia = 1 < 0c 1.1 Again, we define the armature current as E +d - Vt +0 Ia = f jXS

E f +d = Ia (jXS ) + Vt < 0 = 1 (j 0.8) + 1.1+0 1.1+0 = 1 0.8+90c + 1.1+0 1.1+0 = 1.1+0 + 0.8 +90c 1.1 or E f cos d + jE f sin d = 1.1 + j 0.8 1.1 Now, we have the power equation E V P = f t sin d XS E (1.1) 1= f sin d 0.8 E f sin d = 0.8 1.1 So, from equation (i), we get So,

E f cos d = 1.1 Dividing equation (ii) by equation (iii), we get E f sin d 0.8 = 1.1 # 1.1 E f cos d tan d = 0.8 1.21 0.8 d = tan-1 b 1.21 l = 33.47c SOL 3.22

...(i)

...(ii) ...(iii)

Correct answer is 880 rpm. At no load, we have


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Electric Machines

Armature resistance,

ra = 0.5 W

Armature current, Leakage inductance, So, we draw the circuit as

Ia = 2 A L = 0.01 H

Chapter 3

Since, we have 300 V dc shunt motor, so we obtain

nodia E no load (1) = V - Ia Ra E1 = 300 - 2 ^0.5h

or

= 300 - 1 = 299 V Again, when the load is applied, we have Armature resistance,

ra = 0.5 W

Armature current, So, we draw the circuit as

Ia = 15 A

Therefore, we obtain E load (2) = V - Ia Ra E2 = 300 - 15 # 0.5 = 300 - 7.5 = 292.5

or Since, we know that

E\N So, or or SOL 3.23

E1 = N1 N2 E2 299 = 900 292.5 N2

(given N1 = 900 rpm )

N2 = 880 rpm

Correct answer is 23.618 ohm. We have the transformer circuit,


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Electric Machines

Page 169

Given the power absorbed by load; P2 = 4 kW = 4000 W . and the power factor is cos f 2 = 0.89 lagging From the given circuit, voltage across load is V2 = 110 V So, the current through the load is P2 + - f I2 = 2 V2 cos f 2 4000 + - cos-1 0.89 = 110 # 0.89

nodia

= 40.858+ - 27.13c This is the current through secondary winding of transformer. So, the current through primary winding is given by I1 = N 2 I 2 N1 where N1 = number of turns in primary winding N2 = Number of turns in secondary winding. From the given circuit, we have N1 = 2 1 N2 I1 = 1 So, 2 I2 Therefore, I1 = I2 = 40.858+ - 27.13c 2 2

= 20.43+27.13c Since, it is required to improve the input power factor to unity. So, the input current should be red. Therefore, the imaginary part of I1 should pass through reactance X . Thus, current through X is IX = Im {I1} = 20.43 sin (27.13c) = 9.31 Hence, the reactance is X = 220 = 220 9.31 IX = 23.618 W SOL 3.24

Correct option is (B). For long transmission lines (over 200 km), parameters of a line are not lumped but distributed uniformally throughout its length. The ABCD parameter of equivalent p circuit are given as V R S 1 + YZ Z b1 + YZ lW A B 6 W 2 S ...(i) >C DH = S YZ W YZ 1+ b l Y 1 + 6 S 2 W X T The p -equivalent circuit


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Electric Machines

Chapter 3

ABCD parameter of above circuit A B 1 + Y2 Z Zl > H => H C D Y1 + Y2 + Y1 Y2 Z 1 + Y1 Z comparing equation (i) and (ii), we get YZ Zl = B = Z b1 + 6 l where Z = zl = total series impedance/phase Y = yl = total shunt admittance/phase So, Z = j 0.5 # 400 = j200 W Y = j 5 # 10-6 # 400 = j2 # 10-3 W j 2 20-3 # j 200 So, Zl = j 200 c1 + # m = j 200 (- 0.066) 6

...(ii)

nodia = j 200 # 0.934 = j 186.8 W | Zl | = 186.8 W Primary leakage reactance

Ll1 = L1 - 2M = 45 - 2 (20) = 5 mH Self inductance of secondary winding refers to primary side N1 2 V1 2 Ll2 = b N2 l L2 = bV2 l L 2 220 2 = b 110 l # 30

So, Ll2 = 120 mH Leakage inductance of secondary referred to primary side Ll2 = Ll2 - 2M = 120 - 2 (20) = 80 mH Magnetizing inductance, Lm = 2M = 2 (20) = 40 mm SOL 3.25

Correct option is (D). Leakage flux in an induction motor is flux that linked the stator winding or the rotor winding but not both.

SOL 3.26

Correct option is (D). The angle d in the swing equation of a synchronous generator is the angular displacement of an axis fixed to the rotor with respect to a synchronously rotating axis.

SOL 3.27

Correct option is (C). Since, the no-load loss (when load is not connected) is 64 W. So, the open circuit loss is W0 = 6 watt (no load) As the 90% of rated current flows in its both LV and HV windings, the lost is


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Electric Machines

Page 171

measured as copper loss at 90% load = 81 watt or, Wcu ât 90%h = 81 watt Now, we have the copper loss for x load as 2 Wcu ^x h = Wcu ât full loadh # c Ix m IFL 2 I or, Wcu ât full loadh = Wcu ât x% loadh # c FL m I x

= ^81h #

1 ^0.9h2 = 100 watt For maximum efficiency we must have copper lossed = no-load loss = 64 watt Consider at x% load copper loss is 64 watt. So, ^x2h Wcu ât full loadh = 64 64 = 0.8 or, x = 100

nodia x = 80% load

or

SOL 3.28

Correct option is (B). Given,

f = 50 Hz

frequency of source, no. of poles

P =4

rotating speed N = 1440 rpm Now, the synchronous speed is determined as 120f NS = P = 120 ^50h = 1500 rpm 4 So, the slip in the motor is S = NS - N = 1500 - 1440 = 0.04 1500 NS Now, the electrical frequency of the induced negative sequence current in rotor is obtained as

f0 = ^2 - 5h f where f is stator frequency given as f = 50 Hz . Therefore, f0 = ^2 - 0.04h 50 = 98 Hz SOL 3.29

Correct option is (C). For the given transformer, we have V = 1.25 1 VWX


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Electric Machines

Chapter 3

VYZ = 0.8 (attenuation factor) V VYZ = 0.8 1.25 = 1 ^ h^ h VWX

Since, So,

VYZ = VWX

or,

VYZ = 100 100 VWX V = 100 V ; WX = 100 100 VYZ

VWX = 100 V ;

at

1

1

nodia 1

at

VWZ

2

2

2

SOL 3.30


S = ns - n ns

Slip is given as where,

ns = synchronous speed n = rotor speed Thus, slip depend on synchronous speed and the rotor speed. Also, torque increases with increasing slip up to a maximum value and then decreases. Slip does not depend on core/loss component.

SOL 3.31

Correct option is (D). E \ nf where n " speed, f " flux and E " back emf Given that, Vt = 250 V , Ra = 0.25 W Armature current,

Now, Armature current,

n1 = 1000 rpm , IL1 = 68 A , IF1 = 2.2 A Ia1 = IL1 - IF1 = 68 - 2.2 = 65.8 A E1 = Vt - Ia, Ra = 250 - (65.8) (0.25) = 203.55 V n2 = 1600 rpm , IL2 = 52.8 A , IF 2 = 1.8 A Ia = IL2 - IF 2 = 52.8 - 1.8 = 51 A E2 = Vt - Ia Ra 2

2

E1 E2 203.55 207.45

= 220 - (51) (0.25) = 207.25 V f = a n1 kc 1 m n 2 f2 f = b 1000 lc 1 m 1600 f2

f 2 = 0.6369f1 f1 - f2 100 f1 # f - 0.6369f1 = 1 # 100 f1

% reduce in flux =


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Electric Machines

Page 173

- 36.3% SOL 3.32

Correct option is (B). Given that magnetizing current and losses are to be neglected. Locked rotor line current. E2 (R2 = 0) I2 = E2 = 2 Z2 R 2 + X 22 I2 = E2 = E2 \ E2 X2 wL 2 f 50 230 57 So = I l b 236 lb 50 l 2

I2l = 45.0 A SOL 3.33

nodia

Correct option is (B). Since the core length of the second transformer is 2 times of the first, so the core area of the second transformer is twice of the first. Let subscript 1 is used for first transformer and 2 is used for second transform. Area

a2 = 2a1

Length

l2 =

2 l1 N 2 ma L = l

Magnetizing inductance,

N = no. of turns m = length of flux path a = cross section area l = length L\a l

N and m are same for both the transformer

L1 = a1 : l 2 a 2 l1 L2 L 1 = a1 : 2a1 L2

2 l1 l1

L 2 = 2 L1 Thus, magnetizing reactance of second transformer is Magnetizing current

2 times of first.

Xm2 =

2 Xm1 Im = V Xm

Im1 = V1 : Xm2 = V1 2 Xm1 b 2V1 lc Xm1 m Im2 V2 Xm1

(V2 = 2V1)

Im2 = 2 Im1 Thus, magnetizing current of second transformer Im2 = 2 # 0.5 = 0.707 A Since voltage of second transformer is twice that of first and current is that of first, so power will be 2 2 times of first transformer.

2 times

P2 = 2 2 # 55 = 155.6 W SOL 3.34

Correct option is (A). The armature current of DC shunt motor


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Electric Machines

Chapter 3

Ia = V - Eb Ra at the time of starting, Eb = 0 . If the full supply voltage is applied to the motor, it will draw a large current due to low armature resistance. A variable resistance should be connected in series with the armature resistance to limit the starting current. A 4-point starter is used to start and control speed of a dc shut motor. SOL 3.35

SOL 3.36

SOL 3.37

Correct option is (B). The Back emf will go to zero when field is reduced, so Current input will be increased. But when Field increases (though in reverse direction) the back emf will cause the current to reduce.

nodia

Correct option is (C). An air-core transformer has linear B -H characteristics, which implies that magnetizing current characteristic will be perfectly sinusoidal. Correct option is (A).

Initially Eb = V - Ia Ra = 220 - 1 # 10 = 210 V Now the flux is reduced by 10% keeping the torque to be constant, so the current will be 1

Ia f1 = Ia f2 f Ia = Ia 1 = 10 # 1 = 11.11 A 0.9 f2 1

2

2

&

1

Eb \ Nf Nf Eb = 2 2 = 0.9 Eb N1 f 1 2

` f2 = 0.9f1

N1 = N 2

1

Eb = 0.9Eb = 0.9 # 210 = 189 V Now adding a series resistor R in the armature resistor, we have 2

1

Eb = V - Ia (Ra + R) 2

2

189 = 220 - 11.11 (1 + R) R = 1.79 W SOL 3.38

Option ( ) is correct. The steady state speed of magnetic field ns = 120 # 50 = 1000 rpm 6 S = 0.05 nr = (1 - S) ns = 0.95 # 1000 = 950 rpm In the steady state both the rotor and stator magnetic fields rotate in synchronism , so the speed of rotor field with respect to stator field would be zero. Speed of rotor which respect to stator field

Speed of rotor

= nr - ns = 950 - 1000 =- 50 rpm None of the option matches the correct answer. SOL 3.39

Correct option is (C). Given


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Electric Machines

Page 175

I 0 = 1 amp (magnetizing current) Primary current IP = ? I2 = 1 A I2l = secondary current reffered to Primary = 2 # 1 = 2 amp 1

nodia IP =

i 02 + i 22

= 1+4 =

SOL 3.40

SOL 3.41

5 = 2.24 Amp

Option ( ) is correct.

Correct option is (C). Synchronize speed of induction machine 120f = 120 # 50 = 1500 rpm Ns = 4 P Speed of machine = 1600 rpm

= Actual speed of induction machine slip = 1500 - 1600 = - 1 =- 0.066 (negative) 15 1500 Hence induction machine acts as induction generator and dc machine as dc motor. SOL 3.42

Correct option is (B). T = 2.5 Nm at 1400 rpm V =? T = Eb Ib ~ 2.5 = 186.6 # Ia # 60 2p # 1400

than

Ia = 1.963 A V = Eb + Ia Ra = 186.6 + 1.963 # 3.4 = 193.34 V SOL 3.43


SOL 3.44

Correct option is (B). Given no-load speed N1 = 1500 rpm Va = 200 V, T = 5 Nm, N = 1400 rpm emf at no load


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Electric Machines

Chapter 3

Eb1 = Va = 200 V E N \ Eb & N1 = b N2 Eb Eb = b N2 l Eb = 1400 # 200 = 186.67 V 1500 N1 T = Eb Îa /wh & 186.67 # 60 Ia = 5 2p # 1400 1

2

2

1

Ia = 3.926 A V = Eb + Ia Ra Ra = Va - Eb = 200 - 186.67 = 3.4 W 3.926 Ia SOL 3.45


nodia So,

Given field excitation of = 20 A Armature current = 400 A Short circuit and terminal voltage = 200 V On open circuit, load current = 200 A Internal resistance = 2000 = 5 W 400 Internal vol. drop = 5 # 200 = 1000 V

SOL 3.46

SOL 3.47

Correct option is (C). Given single-phase iron core transformer has both the vertical arms of cross section area 20 cm2 , and both the horizontal arms of cross section are 10 cm2 So, Inductance = NBA (proportional to cross section area) 1 When cross section became half, inductance became half. Correct option is (A). Given 3-phase squirrel cage induction motor.

At point A if speed -, Torque speed ., Torque . So A is stable. At point B if speed - Load torque . So B is un-stable. SOL 3.48


SOL 3.49



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Electric Machines

Page 177

Peak voltage across A and B with S open is V = m di = m # (slope of I - t) dt = 400 # 10- 3 # : 10 - 3 D = 800 V p p 5 # 10 SOL 3.50


SOL 3.51

Correct option is (A). Wave form

SOL 3.52

SOL 3.53

SOL 3.54

VAlBl

nodia

Correct option is (B). When both S1 and S2 open, star connection consists 3rd harmonics in line current due to hysteresis A saturation. Correct option is (A). Since S2 closed and S1 open, so it will be open delta connection and output will be sinusoidal at fundamental frequency. Correct option is (A).

N1 N2 I V

= 4000 = 6000 = 25 A = 400 V , f = 50 Hz

Coil are to be connected to obtain a single Phase, 400 V auto transfer to drive 1000 Load 10 kVA. Connected A & D common B


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Electric Machines

Chapter 3

Correct option is (D). Given 3-phase, 400 V, 5 kW, Star connected synchronous motor. Internal Resistance = 10 W Operating at 50% Load, unity p.f. So kVA rating = 25 # 400 = 1000 Internal Resistance = 10 W So kVA rating = 1000 # 10 = 10000 kVA

SOL 3.56

SOL 3.57

SOL 3.58

Correct option is (D). Distributed winding and short chording employed in AC machine will result in reduction of emf and harmonics.

nodia Correct option is (B). Transformer connection will be represented by Y d1.

Correct option is (D). Detent torque/Restraining toque: The residual magnetism in the permanent magnetic material produced. The detent torque is defined as the maximum load torque that can be applied to the shaft of an unexcited motor without causing continuous rotation. In case the motor is unexcited.

SOL 3.59

Correct option is (D). Given: 1-f transformer, 230 V/115 V, 2 kVA W1 : 250 V, 10 A, Low Power Factor W2 : 250 V, 5 A, Low Power Factor W3 : 150 V, 10 A, High Power Factor W4 : 150 V, 5 A, High Power Factor In one circuit test the wattmeter W2 is used and in short circuit test of transformer W3 is used.

SOL 3.60

Correct option is (B). Given: 230 V, 50 Hz, 4-Pole, 1-f induction motor is rotating in clock-wise(forward) direction Ns = 1425 rpm Rotar resistance at stand still(R2 ) = 7.8 W So Ns = 120 # 50 = 1500 4 Slip(S ) = 1500 - 1425 = 0.05 1500 Resistance in backward branch rb = R2 = 7.8 =4W 2 - 0.05 2-S

SOL 3.61

Correct option is (C). Given: a 400 V, 50 Hz, 30 hp, 3-f induction motor Current = 50 A at 0.8 p.f. lagging


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Electric Machines

Page 179

Stator and rotor copper losses are 1.5 kW and 900 W fraction and windage losses = 1050 W Core losses = 1200 W = 1.2 kW So, Input power in stator =

3 # 400 # 50 # 0.8 = 27.71 kW

Air gap power = 27.71 - 1.5 - 1.2 = 25.01 kW SOL 3.62

Correct option is (A). Induced emf in secondary

=- N2

df dt

nodia During - 0 < t < 1,

E1 =- (100)

df =- 12 V dt

E1 and E2 are in opposition

E2 = 2E1 = 24 V df During time 1 < t < 2 , = 0 , then E1 = E2 = 0 dt df During 2 < t < 2.5 , =- 24 V E1 =- (100) dt E2 =- 0 - 48 V

Then

SOL 3.63

Correct option is (A). Given 3-f, 440 V, 50 Hz, 4-Pole slip ring motor Motor is coupled to 220 V

N = 1410 rpm, W1 = 1800 W, W2 = 200 W 120f = 120 # 50 = 1500 rpm Ns = 4 P

So,

Relative speed = 1500 - 1410 = 90 rpm in the direction of rotation. SOL 3.64

Correct option is (C). Neglecting losses of both machines Slip(S ) = Ns - N = 1500 - 1410 = 0.06 1500 Ns total power input to induction motor is Pin = 1800 - 200 = 1600 W Output power of induction motor Pout = (1 - S) Pin = (1 - 0.06) 1600 = 1504 W Losses are neglected so dc generator input power = output power = 1504 W So,

SOL 3.65

2

I R = 1504 I = 1504 = 12.26 A 10

Correct option is (B). Given 400 V, 50 Hz, 4-Pole, 1400 rpm star connected squirrel cage induction


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Electric Machines

Chapter 3

motor. R = 1.00 W, Xs = Xlr = 1.5 W Rlr So, for max. torque slip Sm = Xsm + Xlrm For starting torque Sm = 1 Then Xsm + Xlrm = Rlr 2pfm Ls + 0.2pfm Llr = 1 Frequency at max. torque 1 fm = 2p (Ls + Llr ) Xs = 1.5 Ls = 2p # 50 2p # 50 Llr = 1.5 2p # 50 1 fm = = 50 = 16.7 Hz 1.5 + 1.5 3 50 50 In const V/f control method

nodia V1 = 400 = 8 50 f1 V2 = 8 f1

So

SOL 3.66

V2 = f2 # 8 = 16.7 # 8 = 133.3 V

Correct option is (D). Given: V = 240 V , dc shunt motor I = 15 A Rated load at a speed = 80 rad/s Armature Resistance = 0.5 W Field winding Resistance = 80 W So,

SOL 3.67

E = 240 - 12 # 0.5 = 234 Vplugging = V + E = 240 + 234 = 474 V

Correct option is (A). External Resistance to be added in the armature circuit to limit the armature current to 125%. 474 So Ia = 12 # 1.25 = Ra + R external Ra + R external = 31.6 R external = 31.1 W

SOL 3.68

Correct option is (D). A synchronous motor is connected to an infinite bus at 1.0 p.u. voltage and 0.6 p.u. current at unity power factor. Reactance is 1.0 p.u. and resistance is negligible. So,


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Electric Machines

Page 181

V = 1+0c p.u. Ia = 0.6+0c p.u. Zs = Ra + jXs = 0 + j1 = 1+90c p.u. V = E+d + Ia Zs = 1+0c - 0.6+0c # 1+90c E+d = 1.166+ - 30.96c p.u. Excitation voltage = 1.17 p.u. Load angle (d) = 30.96c(lagging) SOL 3.69


SOL 3.70

Correct option is (C). In transformer zero voltage regulation at full load gives leading power factor.

SOL 3.71

nodia

Correct option is (B). Speed-armature current characteristic of a dc motor is shown as following

The shunt motor provides speed regulation at full load without any controller. SOL 3.72

Correct option is (C). From the given characteristics point A and D are stable

SOL 3.73

Correct option is (B). When the 3-f synchronous motor running at full load and unity power factor and shaft load is reduced half but field current is constant then it gives leading power factor.

SOL 3.74

Correct option is (A). Given star connected synchronous machine, P = 100 kVA Open circuit voltage V = 415 V and field current is 15 A, short circuit armature current at a field current of 10 A is equal to rated armature current. So, Line synchronous impedance open circuit line voltage = 3 # short ckt phase current 415 = = 1.722 100 # 1000 3 #c m 3 # 415

SOL 3.75



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Electric Machines

Chapter 3

Given 1-f transformer P = 50 kVA , V = 250 V/500 V Two winding transformer efficiency 95% at full load unity power factor.

50 # 1 # 1 50 # Wcu + Wi So Wcu + Wi = 2.631 Reconfigured as a 500 V/750 V auto-transformer Efficiency

95% =

nodia auto-transformer efficiency

h=

150 = 98.276% 150 + 2.631

SOL 3.76

Correct option is (B). Given 3-f, 3-stack Variable reluctance step motor has 20-poles Step angle = 360 = 6c 3 # 20

SOL 3.77

Correct option is (D). Given a 3-f squirrel cage induction motor starting torque is 150% and maximum torque 300% So TStart = 1.5TFL Tmax TStart Then Tmax TStart Tmax From equation (1) and (2) 2S max 2 S max + 1 2 S max - 4S max + 1 So S max

SOL 3.78

= 3TFL =1 2 = 22S max 2 S max + 1

...(1) ...(2)

=1 2 =0 = 26.786%

Correct option is (C). Given 3-f squirrel cage induction motor has a starting current of seven the full load current and full load slip is 5%


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Electric Machines

Page 183

ISt = 7I Fl S Fl = 5% TSt = ISt 2 x2 S bTFl l # # Fl TFl 1.5 = (7) 2 # x2 # 0.05 x = 78.252% SOL 3.79

SOL 3.80

Correct option is (B). Star delta starter is used to start this induction motor. So ISt 2 S = 1 72 0.05 TSt = 1 # b 3 3# # TFl I Fl l # Fl TSt = 0.816 TFl

nodia Correct option is (C). Given starting torque is 0.5 p.u. TSt = Isc 2 S So, b I Fl l # Fl TFl 2 0.5 = b Isc l # 0.05 I Fl Per unit starting current Isc = 0.5 = 3.16 A 0.05 I Fl

SOL 3.81

Correct option is (D). In transformer, in open circuit test, current is drawn at low power factor but in short circuit test current drawn at high power factor.

SOL 3.82

Correct option is (B). A single-phase capacitor start induction motor. It has cage rotor and its stator has two windings.

The two windings are displaced 90c in space. The direction of rotation can be changed by reversing the main winding terminals. SOL 3.83

Correct option is (B). In DC motor, compensating winding is used for neutralizing armature reactance while interpole winding is used for improving commutation. Interpoles generate voltage necessary to neutralize the e.m.f of self induction in


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Electric Machines

Chapter 3

the armature coils undergoing commutation. Interpoles have a polarity opposite to that of main pole in the direction of rotation of armature. SOL 3.84

Correct option is (A). Given: A 230 V, DC machine, 20 A at 200 V as a generator. Ra = 0.2 W The machine operated as a motor at same terminal voltage and current, flux increased by 10% So for generator Eg = V + Ia Ra

nodia for motor

So

SOL 3.85

SOL 3.86

Correct option is (B). A synchronous generator is feeding a zero power factor(lagging) load at rated current then the armature reaction is demagnetizing. Correct option is (B). Given the rating of first transformer is 500 kVA Per unit leakage impedance is 0.05 p.u. Rating of second transformer is 250 kVA actual impedance So, Per unit impedance = base impedance and, Per unit leakage impedance \ 1 kVA Then

SOL 3.87

= 200 + 20 # 0.2 Eg = 204 volt Em = V - Ia Ra = 200 - 20 # 0.2 Em = 196 volt N Eg f = g # g Nm Em fm 204 = Ng 1 196 Nm # 1.1 Nm = 196 = 0.87 204 # 1.1 Ng

500 kVA # 0.05 = 250 kVA # x x = 500 # 0.05 = 0.1 p.u. 250

Correct option is (C). Given speed of a 4-pole induction motor is controlled by varying the supply frequency when the ratio of supply voltage and frequency is constant. f = 50 Hz , V = 400 V , N = 1440 rpm So V \f V1 = f1 V2 f2 V2 = 400 # 30 = 240 V 50


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Electric Machines

So

Page 185

2 T \ cV m S f S2 = V1 2 f2 T2 S1 bV2 l # f1 # T1

Given

T1 = T2

Then

2 S2 = 0.04 # b 400 l # 30 50 240

S2 = 0.066 Nr = Ns (1 - S) 120f P Nr = 120 # 30 ^1 - 0.066h 4 Nr =

So

nodia = 840.6 rpm

SOL 3.88

Correct option is (A). Given a 3-f induction motor

P = 4 , V = 400 V , f = 50 Hz r1 = 1.0 W , r2l= 0.5 W X1 = Xl2 = 1.2 W, Xm = 35 W So, Speed of motor is 120f Ns = = 120 # 50 = 1500 rpm 4 P Torque

SOL 3.89

V2 rl2 Tst = 180 # 2pNs (r1 + rl2) 2 + X 2 400 2 0.5 c 3m# 180 = 2 # 3.14 # 1500 # (1.5) 2 + (2.4) 2 = 63.58 Nm

Correct option is (B). Given that 3-f induction motor star connected P = 10 kW , V = 400 V, Poles = 4, f = 50 Hz Full load current I Fl = 20 A output Efficiency = input So Cu losses at full load 2 = b 20 l # 762 = 1354.67 15 Total losses = 1354.67 + 1002 = 2356.67 10000 Efficiency = 100 = 81% 10000 + 2356.67 #

SOL 3.90

Correct option is (A). Given 3-f star connected synchronous motor internal reactance = 10 W


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Electric Machines

Chapter 3

Operating at 50% load, unity power factor, 400 V, 5 kW Excitation increased = 1% So, full load current I Fl =

5 # 103 = 7.22 3 # 400 # 1

E 2 = (V cos q - Ia Ra) 2 + (V sin q - Ia Xs) 2 400 2 + 10 3.6 2 = 2133.7289 ^ # h 3m Excitation will increase 1% then E2

So,

E =

c

E2 = 2133.7289 # 0.01 = 236 Ia X =

(E2) 2 - V 2

nodia (236) 2 - c 400 m = 48.932 3 Ia = 48.932 = 4.8932 10 4 . 8932 Load (%) = = 67.83 % 7.22 =

SOL 3.91

2

Correct option is (C). Given P = 4 , f = 50 Hz Slots = 48 , each coil has 10 turns Short pitched by an angle(a) to 36c electrical Flux per pole = 0.05 Wb So,

E ph = 4.44 ffTph KW Slot/Pole/ph = 48 = 4 4#3 Slot/Pole = 48 = 12 4 Slot angle = 180 = 15c 12 sin (4 # 15/2) Kd = = 0.957 4 sin (15/2) K p = cos a = cos 18c = 0.951 2 In double layer wdg No. of coil = No of slots No. of turns/ph = 48 # 10 = 160 3 Then

E ph = 4.44 # 0.025 # 50 # 0.957 # 0.951 # 160 = 808 V EL =

3 # 808 EL = 1400 V (approximate) SOL 3.92

Correct option is (A). line to line induced voltage, so in 2 phase winding Slot/ pole/ph = 6


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Electric Machines

Page 187

Tph = 480 = 240 2 Slot angle = 180 # 4 = 15c 48 sin 6 # (15/2) Kd = = 0.903 6 sin (15/2) K p = cos b 36 l = 0.951 2 E ph = 4.44 # 0.025 # 50 # 240 # 0.951 # 0.903 = 1143 SOL 3.93

SOL 3.94

Correct option is (A). Fifth harmonic component of phase emf So Angle = 180 = 36c 5 the phase emf of fifth harmonic is zero.

nodia Correct option is (C). Given that: A 300 kVA transformer Efficiency at full load is 95% and 0.8 p.f. lagging 96% efficiency at half load and unity power factor So For Ist condition for full load kVA # 0.8 95% = kVA # 0.8 + Wcu + Wi

...(1)

Second unity power factor half load 96% = So Then SOL 3.95

...(2)

Wcu + Wi = 12.63 0.25Wcu + 0.96Wi = 6.25 Wcu = 8.51, Wi = 4.118

Correct option is (B). Efficiency (h) = So

SOL 3.96

kVA # 0.5 kVA # 0.5 + Wcu + Wi

X # p.f. # kVA X # kVA + Wi + Wcu # X2

4.118 = 0.6956 8.51 0.6956 # 1 # 300 h% = 0.6956 # 300 + 4.118 + 8.51 # (0.6956) 2 h = 96.20% X =

Correct option is (D). The leakage reactances X1 , and X2l are equal and magnetizing reactance Xm is higher than X1 , and X2l X1 . X2l << Xm


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Electric Machines

Chapter 3

SOL 3.97

Correct option is (B). Three phase star delta connection of transformer induces a phase difference of 30c between output and input line voltage.

SOL 3.98

Correct option is (A). Given torque/speed curve of the induction motor

nodia

When the speed of the motor is in forward direction then slip varies from 0 to 1 but when speed of motor is in reverse direction or negative then slip is greater then 1. So at point W slip is greater than 1.

SOL 3.99

Correct option is (B). For an induction motor the ratio of gross power output to air-gap is equal to (1 - s) gross power So = (1 - s) airgap power

SOL 3.100

Correct option is (A). Given that two magnetic pole revolve around a stationary armature. At c1 the emf induced upward and no emf induced at c2 and c2l

SOL 3.101

Correct option is (B). Given A 50 kW DC shunt motor is loaded, then at half the rated speed by armature voltage control So P\N

Pnew = 50 = 25 kW 2 At 1.5 time the rated speed by field control P = constant P = 50 kW

So SOL 3.102



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Electric Machines

Page 189

In synchronous machine, when the armature terminal are shorted the field current should first be decreased to zero before started the alternator. In open circuit the synchronous machine runs at rated synchronous speed. The field current is gradually increased in steps. The short circuit ratio is the ratio of field current required to produced the rated voltage on open to the rated armature current. SOL 3.103

Correct option is (D). E = PNf b Z l A

In DC motor,

or E = Kfwn So armature emf E depends upon f and w only and torque developed depends upon PZfIa T = 2pA So, torque(T ) is depends of f and Ia and developed power(P ) is depend of flux f, speed w and armature current Ia .

SOL 3.104

SOL 3.105

nodia Correct option is ( ).

Correct option is (B). Given a three-phase cage induction motor is started by direct on line switching at rated voltage. The starting current drawn is 6 time the full load current. Full load slip = 4% So ISt 2 TSt bTFl l = b I Fl l # S Fl = (6) 2 # 0.04 = 1.44

SOL 3.106

Correct option is (B). Given single-phase induction motor driving a fan load, the resistance rotor is high So ...(1) Eb = V - Ia Ra Pmech = Ea Ia t = Pmech wm

...(2)

From equation (1) and (2) the high resistance of rotor then the motor achieves quick acceleration and torque of starting is increase. SOL 3.107

Correct option is (A). Given V/f control of induction motor, the maximum developed torque remains same we have, E = 4.44Kw ffT1 If the stator voltage drop is neglected the terminal voltage E1 . To avoid saturation and to minimize losses motor is operated at rated airgap flux by varying terminal voltage with frequency. So as to maintain (V/f ) ratio constant at the rated 1


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Electric Machines

Chapter 3

value, the magnetic flux is maintained almost constant at the rated value which keeps maximum torque constant. SOL 3.108

Correct option is (B). Given P = 1000 kVA , 6.6 kV Reactance = 20 W and neglecting the armature resistance at full load and unity power factor P =

So

I =

3 VL IL 1000 = 87.47 A 3 # 6. 6

nodia IX = 87.47 # 20 = 1.75 kV 2 2 = c 6.5 m + (1.75) 2 E ph 3

So,

6.5 2 2 c 3 m + (1.75) = 4.2 kV

E ph = E ph

EL = 3 E ph EL = 1.732 # 4.2 = 7.26 kV SOL 3.109

a Star connection

Correct option is (C). Torque angle az = tan- 1 b Xs l Ra

az = tan- 1 c SOL 3.110

3 # 1.75 = 24.6c m 6.6

Correct option is (B). Given that Transformer rating is 500 kVA Iron losses = 300 W full load copper losses = 600 W Maximum efficiency condition Wi = X2 Wc


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Electric Machines

So,

X =

Wi = Wc

Page 191

300 = 0.707 600

efficiency% = 0.707 # 100 = 70.7% SOL 3.111

Correct option is (C). Stepper motor is rotated in steps, when the supply is connected then the torque is produced in it. The higher value of torque is pull out torque and less torque when the torque is pull in torque.

SOL 3.112

Correct option is (C). The stepper motor has the permanent magnet rotor and stator has made of windings, it’s connected to the supply.

SOL 3.113

SOL 3.114

SOL 3.115

nodia

Correct option is (D). 1-phase induction motor is not self starting, so it’s used to start different method at full load condition, capacitor-run type motor have higher power factor. In this type the capacitor is connected in running condition. Correct option is (C). Given that if 3-f induction motor is rotated in clockwise then the phase sequence of supply voltage is A-B-C. In counter clock wise rotation of the motor the phase sequence is change so in the counter clockwise rotation the phase sequence is A-C-B. Correct option is (A). In linear electromagnetic co-energy.

circuit

the

field

energy

is

equal

to

the

Wf = W f' = 1 Li2 = 1 yi = 1 y2 2 2 2L Wf = field energy W f' = co energy

SOL 3.116

Correct option is (A). Given that 8-Pole, 50 Hz induction machine in seventh space harmonic mmf wave. So, Synchronous speed at 7th harmonic is = Ns /7 120f Speed of motor Ns = = 120 # 50 = 750 rpm 8 P


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Electric Machines

Chapter 3

Synchronous speed is = Ns = 750 = 107.14 rpm in forward direction 7 7 SOL 3.117

Correct option is (B). Rotating electrical machines having its self inductance of stator and rotor windings is independent of the rotor position of synchronizing torque. synchronizing torque Tsynchronizing = 1 m dP Nm/elect. radian ws dd = b 1 m dP l pP Nm/mech.degree ws dd 180

SOL 3.118

Correct option is (A). Given that the armature of a permanent magnet dc motor is

nodia Ra = 0.8 W

At no load condition

V = 25 V , I = 1.5 A , N = 1500 rpm No load losses = E # I a E = V - IRa So no load losses = (25 - 1.5 # 0.8) 1.5 = 35.7 W At load condition I = 3.5 A Iron losses = I2 R = (3.5) 2 # 0.8 = 9.8 W Total losses = No load losses + iron losses Total power P

P P Efficiency h

= 35.7 + 9.8 = 45.5 W = VI = 25 # 3.5 = 87.5 W output = input total power - losses = total power = 87.5 - 45.5 # 100 87.5 = 48.0%

SOL 3.119

Correct option is (D). Given that 50 kVA, 3300/230 V, 1-f transform

Vin = 3300 V Vout = 3300 + 230 = 3530 V Output current I2 and output voltage 230 V


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Electric Machines

Page 193

So 3 I2 = 50 # 10 = 217.4 A 230 When the output voltage is Vout then kVA rating of auto transformer will be

I2 = 3530 # 217.4 = 767.42 kVA SOL 3.120

Correct option is (D). Given that 100 kVA, 11000/400 V, Delta-star distribution transformer resistance is 0.02 pu and reactance is 0.07 pu So pu impedance Z pu = 0.02 + j0.07 Base impedance referred to primary 2 (11 # 103) 2 = 3630 W Z Base = V P = VL IL /3 100 # 103 3 The phase impedance referred to primary

nodia Z primary = Z pu # Z Base = (0.02 + j0.07) (3630) = 72.6 + j254.1

SOL 3.121

Correct option is (A). Given that 230 V, 50 Hz, 4-Pole, capacitor-start induction motor

Zm = Rm + Xm = 6.0 + j4.0 W ZA = RA + XA = 8.0 + j6.0 W Phase angle of main winding +I m = + - Z m =- + (6 + j4) =- +33.7c So angle of the auxiliary winding when the capacitor is in series. +IA =- + (8 + j6) + 1 jwC j = + (8 + j6) wC a = +I A - +I m So

90 =- tan >f -1

6- 1 wC p - (- 33.7)H 8

1 = 18 wC

w = 2pf


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Electric Machines

So

C =

Chapter 3

1 1 = 18 # 2 # 3.14 # 50 18 # 2pf

= 176.8 mF SOL 3.122

Correct option is (A). Given that the armature has per phase 1.7 W and two alternator is connected in parallel So,

synchronous

reactance

of

nodia both alternator voltage are in phase So, E f1 = 3300 3 E f2 = 3200 3 Synchronizing current or circulating current EC = TS1 + TS2 Reactance of both alternator are same E - Ef 2 So = f1 TS 1 + TS 2 = 1 b 3300 - 3200 l = 16.98 A 3 1.7 + 1.7

SOL 3.123

Correct option is (C). Given V = 400 V, 15 kW power and P = 4 f = 50 Hz , Full load slip (S ) = 4% 120f So = 120 # 50 = 1500 rpm Ns = 4 P Actual speed = synchronous speed - slip N = 1500 - 4 # 1500 = 1440 rpm 100 P Torque developed , where ws (1 - S) = 2pN T = 60 ws (1 - S) 3 = 15 # 10 # 60 = 99.47 Nm 2p # 1440

SOL 3.124

Correct option is (B). Given 1.8c angle, 2-f Bipolar stepper motor and stepping rate is 100 step/second So step required for one revolution = 360 = 200 steps 1.8 a Time required for one revolution = 2 seconds rev/sec = 0.5 rps


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Electric Machines

and SOL 3.125

Page 195

rev/min = 30 rpm

Correct option is (C). Given that: P = 8 Pole, DC generator has wave-wound armature containing 32 coil of 6 turns each. Simplex wave wound flux per pole is 0.06 Wb N = 250 rpm So induced armature voltage fZNP Eg = 60A Z = total no.of armature conductor = 2CNC = 2 # 32 # 6 = 384 Eg = 0.06 # 250 # 3.84 # 8 60 # 2

nodia

a A = 2 for wave winding

Eg = 384 volt

SOL 3.126

Correct option is (C). Given a 400 V, 50 Hz and 0.8 p.f. loading delta connection 50 Hz synchronous machine, the reactance is 2 W. The friction and windage losses are 2 kW and core losses is 0.8 kW and shaft is supply 9 kW at a 0.8 loading power factor So Input power = 9 kW + 2 kW + 0.8 kW = 11.8 kW

a

Input power = I2 =

SOL 3.127

3 V2 I2 = 11.8 kW 11.8 kW = 21.29 A 3 # 400 # 0.8

Correct option is (B). Given that 500 MW, 3-f star connected synchronous generator has a rated voltage of 21.5 kV and 0.85 Power factor So

3 VL IL = 500 MW 500 # 106 IL = 3 # 21.5 # 103 # 0.85 = 15.79 # 103 IL = 15.79 kA

SOL 3.128

Correct option is (D). Given that 1-f transformer, maximum efficiency 90% at full load and unity power factor V2 I2 cos f2 So h= V2 I2 cos f2 + Pi + Pc (L.F) cos f2 = (L.F) cos f2 + Pi(Pu) + Pc where L.F. is the load fator. At full load, load factor is L.F. = Pi = 1 Pc


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Electric Machines

so,

Chapter 3

cos f2 = 1 at unity power factor 90% = 1 # 1 1 + 2Pi

Pi = 0.0555 MVA At half load, load factor is L.F = 1 = .5 2 0.5 # 1 So, h= # 100 = 87.8% 0.5 # 0.0555 # (0.5) 2 + 0.0555 SOL 3.129

SOL 3.130

SOL 3.131

Correct option is (C). In food mixer the universal motor is used and in cassette tap recorder permanent magnet DC motor is used. The Domestic water pump used the single and three phase induction motor and escalator used the three phase induction motor.

nodia

Correct option is (D). Given a engine drive synchronous generator is feeding a partly inductive load. A capacitor is connected across the load to completely nullify the inductive current. Then the motor field current has to be reduced and fuel input left unaltered. Correct option is (A).

Given open circuit and full-load zero power factor of a synchronous generator. At point Q the zero power factor at 1.0 pu voltage. The voltage drop at point PQ is across synchronous reactance. SOL 3.132

Correct option is (D). Given no load test on 3-f induction motor, the graph between the input power and voltage drop is shown in figure, the intersection point yield the friction and windage loss.

SOL 3.133



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Electric Machines

Page 197


Given that: The armature resistance in per unit is 0.2 so, Ra = 0.2 back emf equation of motor is Eb = V - Ia Ra given that no mechanical losses and armature reaction is neglected, so per unit value of emf induced by motor is

nodia

Eb = 0.98 The DC shunt motor is mechanically coupled by the generator so the emf induced by motor and generator is equal

Eg = Eb so voltage generated by the generator is

V = 0.98 - 1 # 0.2 = 0.96 per unit value of load resistance is equal to 0.96

SOL 3.135

SOL 3.136

Correct option is (D). Given that when the switch S 1 is closed and S 2 is open then the 11000 V is step down at 415 V output Second time when the switch S 1 is open and switch S 2 is closed then 2-phase supply is connected to the transformer then the ratio of voltage is V1 = N1 = 11000 = 26.50 415 V2 N2 The output terminal a and c are in opposite phase so cancelled with each other and terminal is equal to zero volt. Correct option is (D). Given that Resistance So

R V1 V1 V2 V2

and

V1 V3

N1 : N2: N 3 is 4: 2: 1 = 10 W = 400 V = N1 = 4 N2 2 = 2V1 = 200 V 4 = N1 = 4 N3 1

V3 = 100 V so current in secondary winding I2 = V2 = 200 = 20 A 10 R The current in third winding when the capacitor is connected


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Electric Machines

Chapter 3

I 3 = V3 = 100 = j40 - jXc - j2.5 When the secondary winding current I2 is referred to primary side i.e I 1' I 1' = N2 = 2 So I2 N1 4 I 1' = 20 = 10 A 2 and winding third current I 3 is referred to Primary side i.e I 1'' . I 3 flows to opposite to I1 I 1'' = N 3 = 1 So N1 4 - I3 so

I 1'' =- j10 So total current in primary winding is

nodia I1 = I 1'' + I 2'' = 10 - j10 A

SOL 3.137

Correct option is (A). Given that: P Stator winding current is dc, rotor winding current is ac Q R S

Stator winding current is ac, rotor winding current is dc Stator winding current is ac, rotor winding current is ac Stator has salient pole and rotor has commutator

T U

Rotor has salient pole and slip rings and stator is cylindrical Both stator and rotor have poly-phase windings

So DC motor/machines: The stator winding is connected to dc supply and rotor winding flows ac current. Stator is made of salient pole and Commutator is connected to the rotor so rotor winding is supply ac power. Induction machines: In induction motor the ac supply is connected to stator winding and rotor and stator are made of poly-phase windings. Synchronous machines: In this type machines the stator is connected to ac supply but rotor winding is excited by dc supply. The rotor is made of both salient pole and slip rings and stator is made of cylindrical. SOL 3.138

Correct option is (C). Given that Fs is the peak value of stator mmf axis. Fr is the peak value of rotor mmf axis. The rotor mmf lags stator mmf by space angle d. The direction of torque acting on the rotor is clockwise or counter clockwise. When the opposite pole is produced in same half portion of stator and rotor then the rotor moves. So portion of stator is north-pole in ABC and rotor abc is produced south pole as well as portion surface CDA is produced south pole and the rotor cda is produced North pole. The torque direction of the rotor is clock wise and torque at surface is in counter clockwise direction.


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Electric Machines

Page 199

Correct option is (A). Given that: A 4-pole, 3-f, double layer winding has 36 slots stator with 60c phase spread, coil span is 7 short pitched so, Pole pitch = slot = 36 = 9 4 pole Slot/pole/phase = 3 so, 3-slots in one phase, if it is chorded by 2 slots then Out of 3 " 2 have different phase Out of 36 " 24 have different phase. Correct option is (B). Given that: 3-f induction motor is driving a constant load torque at rated voltage and frequency. Voltage and frequency are halved and stator resistance, leakage reactance and core losses are ignored. Then the motor synchronous speed and actual speed difference are same. 120f Ns = P The leakage reactance are ignored then the air gap flux remains same and the stator resistance are ignored then the stator current remain same.

SOL 3.140

nodia

SOL 3.141

Correct option is (D). Given that: 1-f induction motor main winding excited then the rotating field of motor changes, the forward rotating field of motor is greater then the back ward rotating field.

SOL 3.142

Correct option is (B). Given that: A dc series motor driving a constant power load running at rated speed and rated voltage. It’s speed brought down 0.25 pu. Then Emf equation of dc series motor E = V - (Ra + Rse) Ra + Rse = R so, E = V - IR = KfN then N = E Kf In series motor faI so, N = V - IR KI At constant power load E # I = T # W = Const T = KfI = KI 2 If W is decreased then torque increases to maintain power constant. T \ I2 W = 1 then T = 4 4

...(1) ...(2)


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Electric Machines

Chapter 3

So current is increased 2 time and voltage brought down to 0.5 pu. ***********

nodia


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Power Systems

Page 201

CHAPTER 4 POWER SYSTEMS


ONE MARK

Three-phase to ground fault takes place at locations F1 and F2 in the system shown in the figure.

nodia

If the fault takes place at location F1 , then the voltage and the current at bus A are VF1 and IF1 respectively. If the fault takes place at location F2 , then the voltage and the current at bus A and VF2 and IF2 respectively. The correct statement about voltages and currents during faults at F1 and F2 is (B) VF1 leads IF1 and VF2 lags IF2 (A) VF1 leads IF1 and VF2 leads IF2 (C) VF1 lags IF1 and VF2 leads IF2 (D) VF1 lags IF1 and VF2 lags IF2

EE SP 4.2

A 2-bus system and corresponding zero sequence network are shown in the figure.

The transformers T1 and T2 are connected as


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Power Systems

Chapter 4


TWO MARKS

In an unbalanced three phase system, phase current Ia = 1 ^ - 90ch pu , negative sequence current Ib = 4 ^- 150ch pu , zero sequence current Ic = 3 90c pu . The magnitude of phase current Ib in pu is (A) 1.00 (B) 7.81 (C) 11.53 (D) 13.00 2

EE SP 4.4

0

A three phase synchronous generator is to be connected to the infinite bus. The lamps are connected as shown in the figure for the synchronization. The phase sequence of bus voltage is R-Y-B and that of incoming generator voltage is Rl - Yl - Bl.

nodia

It was found that the lamps are becoming dark in the sequence La - Lb - Lc . It means that the phase sequence of incoming generator is (A) opposite to infinite bus and its frequency is more than infinite bus (B) opposite to infinite bus but its frequency is less than infinite bus (C) same as infinite bus and its frequency is more than infinite bus (D) same as infinite bus and its frequency is less than infinite bus EE SP 4.5

A distribution feeder of 1 km length having resistance, but negligible reactance, is fed from both the ends by 400 V, 50 Hz balanced sources. Both voltage sources S1 and S2 are in phase. The feeder supplies concentrated loads of unity power factor as shown in the figure.

The contributions of S1 and S2 in 100 A current supplied at location P respectively, are (A) 75 A and 25 A


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Power Systems

Page 203

(B) 50 A and 50 A (C) 25 A and 75 A (D) 0 A and 100 A EE SP 4.6

A two bus power system shown in the figure supplies load of 1.0 + j0.5 p.u.

nodia The values of V1 in p.u. and d 2 respectively are (A) 0.95 and 6.00c

(B) 1.05 and - 5.44c (C) 1.1 and - 6.00c (D) 1.1 and - 27.12c

EE SP 4.7

The fuel cost functions of two power plants are

C1 = 0.05Pg 12 + APg1 + B C2 = 0.10Pg 22 + 3APg2 + 2B

Plant P1 : Plant P2 :

where, Pg1 and Pg2 are the generated powers of two plants, and A and B are the constants. If the two plants optimally share 1000 MW load at incremental fuel cost of 100 Rs/MWh, the ratio of load shared by plants P1 and P2 is (A) 1 : 4 (B) 2 : 3 (C) 3 : 2 (D) 4 : 1 EE SP 4.8

The over current relays for the line protection and loads connected at the buses are shown in the figure.

The relays are IDMT in nature having the characteristic 0.14 # Time Multiplier Setting ^Plug Setting Multiplierh0.02 - 1 The maximum and minimum fault currents at bus B are 2000 A and 500 A respectively. Assuming the time multiplier setting and plug setting for relay RB to be 0.1 and 5 A respectively, the operating time of RB (in seconds) is_____. top =


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Power Systems

Chapter 4

YEAR 2014 EE02

ONE MARK

EE SP 4.9

A three phase star-connected load is drawing power at a voltage of 0.9 pu and 0.8 power factor lagging. The three phase base power and base current are 100 MVA and 437.38 A respectively. The line-to-line load voltage in kV is _____.

EE SP 4.10

Shunt reactors are sometimes used in high voltage transmission systems to (A) limit the short circuit current through the line (B) compensate for the series reactance of the line under heavily loaded condition (C) limit over-voltages at the load side under lightly loaded condition. (D) compensate for the voltage drop in the line under heavily loaded condition.

nodia YEAR 2014 EE02

EE SP 4.11

EE SP 4.12

TWO MARKS

A synchronous generator is connected to an infinite bus with excitation voltage E f = 1.3 pu . The generator has a synchronous reactance of 1.1 pu and is delivering real power (P) of 0.6 pu to the bus. Assume the infinite bus voltage to be 1.0 pu. Neglect stator resistance. The reactive power (Q) in pu supplied by the generator to the bus under this condition is _____.

There are two generators in a power system. No-load frequencies of the generators are 51.5 Hz and 51 Hz, respectively, and both are having droop constant of 1 Hz/MW . Total load in the system is 2.5 MW. Assuming that the generators are operating under their respective droop characteristics, the frequency of the power system in Hz in the steady state is ______.

EE SP 4.13

The horizontally placed conductors of a single phase line operating at 50 Hz are having outside diameter of 1.6 cm, and the spacing between centers of the conductors is 6 m. The permittivity of free space is 8.854 # 10-12 F/m . The capacitance to ground per kilometer of each line is (B) 8.4 # 10-9 F (A) 4.2 # 10-9 F (C) 4.2 # 10-12 F (D) 8.4 # 10-12 F

EE SP 4.14

A three phase, 100 MVA, 25 kV generator has solidly grounded neutral. The positive, negative, and the zero sequence reactances of the generator are 0.2 pu, 0.2 pu and 0.05 pu, respectively, at the machine base quantities. If a bolted single phase to ground fault occurs at the terminal of the unloaded generator, the fault current in amperes immediately after the fault is_____. YEAR 2014 EE03

EE SP 4.15

ONE MARK

In a long transmission line with r , l , g and c are the resistance, inductance, shunt conductance and capacitance per unit length, respectively, the condition for distortionless transmission is (A) rc = lg (B) r = l/c (C) rg = lc

(D) g = c/l


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Power Systems

Page 205

For a fully transposed transmission line (A) positive, negative and zero sequence impedances are equal (B) positive and negative sequence impedances are equal (C) zero and positive sequence impedances are equal (D) negative and zero sequence impedances are equal

EE SP 4.17

A 183 bus power system has 150 PQ buses and 32 PV buses. In the general case, to obtain the load flow solution using Newton-Raphson method in polar coordinates, the minimum number of simultaneous equations to be solved is _____. YEAR 2014 EE03

EE SP 4.18

EE SP 4.19

EE SP 4.20

TWO MARKS

nodia

For a 400 km long transmission line, the series impedance is ^0.0 + j0.5h W/km and the shunt admittance is ^0.0 + j5.0h mmho/km . The magnitude of the series impedance (in W ) of the equivalent p circuit of the transmission line is ____.

The complex power consumed by a constant-voltage load is given by ^P1 + jQ1h, where, 1 kW # P1 # 1.5 kW and 0.5 kVAR # Q1 # 1 kVAR A compensating shunt capacitor is chosen such that Q # 0.25 kVAR , where Q is the net reactive power consumed by the capacitor load combination. The reactive power (in kVAR) supplied by the capacitor is ____. The figure shows the single line diagram of a single machine infinite bus system.

The inertia constant of the synchronous generator H = 5 MW-s/MVA. Frequency is 50 Hz. Mechanical power is 1 pu. The system is operating at the stable equilibrium point with rotor angle d equal to 30c. A three phase short circuit fault occurs at a certain location on one of the circuits of the double circuit transmission line. During fault, electrical power in pu is Pmax sin d . If the values of d and dd/dt at the instant of fault clearing are 45c and 3.762 radian/s respectively, then Pmax (in pu) is ______. YEAR 2013 EE SP 4.21

TWO MARKS

For a power system network with n nodes, Z 33 of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is 1.3 - 10c per unit. If a capacitor having reactance of - j3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit is (A) 0.325 - 100c (B) 0.325 80c (C) 0.371 - 100c

(D) 0.433 80c


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Power Systems

Chapter 4

Statement for Linked Answer Questions: 22 and 23 In the following network, the voltage magnitudes at all buses are equal to 1 pu, the voltage phase angles are very small, and the line resistances are negligible. All the line reactances are equal to j1 W

EE SP 4.22

EE SP 4.23

nodia The voltage phase angles in rad at buses 2 and 3 are (B) q2 = 0 , q3 =- 0.1 (A) q2 =- 0.1, q3 =- 0.2 (C) q2 = 0.1, q3 = 0.1 (D) q2 = 0.1, q3 = 0.2

If the base impedance and the line-to line base voltage are 100 ohms and 100 kV respectively, then the real power in MW delivered by the generator connected at the slack bus is (A) - 10 (B) 0 (D) 20 (C) 10 YEAR 2012

ONE MARK

EE SP 4.24

The bus admittance matrix of a three-bus three-line system is R- 13 10 5VW S Y = j S 10 - 18 10W SS 5 10 - 13WW T X If each transmission line between the two buses is represented by an equivalent p -network, the magnitude of the shunt susceptance of the line connecting bus 1 and 2 is (B) 2 (A) 4 (C) 1 (D) 0

EE SP 4.25

A two-phase load draws the following phase currents : i1 (t) = Im sin (wt - f 1), i2 (t) = Im cos (wt - f 2). These currents are balanced if f 1 is equal to. (B) f 2 (A) - f 2 (C) (p/2 - f 2) (D) (p/2 + f 2)

EE SP 4.26

The figure shows a two-generator system applying a load of PD = 40 MW , connected at bus 2.


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Power Systems

Page 207

The fuel cost of generators G1 and G2 are : C1 (PG1) = 10000 Rs/MWh and C2 (PG2) = 12500 Rs/MWh and the loss in the line is Ploss (pu) = 0.5PG21 (pu), where the loss coefficient is specified in pu on a 100 MVA base. The most economic power generation schedule in MW is (B) PG1 = 22, PG2 = 20 (A) PG1 = 20, PG2 = 22 (D) PG1 = 0, PG2 = 40 (C) PG1 = 20, PG2 = 20 EE SP 4.27

nodia

The sequence components of the fault current are as follows : I positive = j 1.5 pu, I negative =- j 0.5 pu, I zero =- j1 pu . The type of fault in the system is (A) LG (B) LL (C) LLG (D) LLLG YEAR 2012

EE SP 4.28

For the system below, SD1 and SD2 are complex power demands at bus 1 and bus 2 respectively. If V2 = 1 pu , the VAR rating of the capacitor (QG2) connected at bus 2 is

(A) 0.2 pu (C) 0.312 pu EE SP 4.29

TWO MARKS

(B) 0.268 pu (D) 0.4 pu

A cylinder rotor generator delivers 0.5 pu power in the steady-state to an infinite bus through a transmission line of reactance 0.5 pu. The generator no-load voltage is 1.5 pu and the infinite bus voltage is 1 pu. The inertia constant of the generator is 5 MW- s/MVA and the generator reactance is 1 pu. The critical clearing angle, in degrees, for a three-phase dead short circuit fault at the generator terminal is (A) 53.5 (B) 60.2 (C) 70.8 (D) 79.6 YEAR 2011

EE SP 4.30

ONE MARK

A nuclear power station of 500 MW capacity is located at 300 km away from a load center. Select the most suitable power evacuation transmission configuration among the following options


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EE SP 4.31

EE SP 4.32

Power Systems

Chapter 4

nodia A negative sequence relay is commonly used to protect (A) an alternator (B) an transformer (C) a transmission line (D) a bus bar

For enhancing the power transmission in along EHV transmission line, the most preferred method is to connect a (A) Series inductive compensator in the line (B) Shunt inductive compensator at the receiving end (C) Series capacitive compensator in the line (D) Shunt capacitive compensator at the sending end YEAR 2011

EE SP 4.33

TWO MARKS

A load center of 120 MW derives power from two power stations connected by 220 kV transmission lines of 25 km and 75 km as shown in the figure below. The three generators G1, G2 and G 3 are of 100 MW capacity each and have identical fuel cost characteristics. The minimum loss generation schedule for supplying the 120 MW load is

P1 = 80 MW + losses (A) P2 = 20 MW P3 = 20 MW

P1 = 60 MW (B) P2 = 30 MW + losses P3 = 30 MW


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Power Systems

P1 = 40 MW (C) P2 = 40 MW P3 = 40 MW + losses EE SP 4.34

P1 = 30 MW + losses (D) P2 = 45 MW P3 = 45 MW

The direct axis and quadrature axis reactances of a salient pole alternator are 1.2 p.u and 1.0 p.u respectively. The armature resistance is negligible. If this alternator is delivering rated kVA at upf and at rated voltage then its power angle is (A) 30c (B) 45c (C) 60c

EE SP 4.35

Page 209

(D) 90c

nodia

A three – bus network is shown in the figure below indicating the p.u. impedance of each element.

The bus admittance matrix, Y -bus, of the network is R 0.3 - 0.2 0 V R- 15 0 VW 5 S W S (A) j S- 0.2 0.12 0.08W (B) j S 5 7.5 - 12.5W SS 0 0.08 0.02WW SS 0 - 12.5 2.5 WW XV TR V X RT0.1 0.2 0 0 10 5 S W S W (C) j S0.2 0.12 - 0.08W (D) j S 5 7.5 12.5W SS 0 12.5 - 10WW SS 0 - 0.08 0.10 WW T X T X

Statement For Linked Answer Questions : 36 and 37 EE SP 4.36

Two generator units G1 and G2 are connected by 15 kV line with a bus at the mid-point as shown below

G1 = 250 MVA , 15 kV, positive sequence reactance XG = 25% on its own base G2 = 100 MVA , 15 kV, positive sequence reactance XG = 10% on its own base L1 and L2 = 10 km , positive sequence reactance XL = 0.225 W/km 1

2


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Power Systems

Chapter 4

nodia EE SP 4.37

In the above system, the three-phase fault MVA at the bus 3 is (A) 82.55 MVA (B) 85.11 MVA (C) 170.91 MVA (D) 181.82 MVA YEAR 2010

EE SP 4.38

ONE MARK

Power is transferred from system A to system B by an HVDC link as shown in the figure. If the voltage VAB and VCD are as indicated in figure, and I 2 0 , then


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Power Systems

Page 211

(A) VAB 1 0,VCD < 0,VAB > VCD (B) VAB 2 0,VCD 2 0,VAB 1 VCD (C) VAB 2 0,VCD 2 0,VAB > VCD (D) VAB 2 0,VCD < 0 EE SP 4.39

Consider a step voltage of magnitude 1 pu travelling along a lossless transmission line that terminates in a reactor. The voltage magnitude across the reactor at the instant travelling wave reaches the reactor is

nodia (A) - 1 pu (C) 2 pu

EE SP 4.40

(B) 1 pu

(D) 3 pu

Consider two buses connected by an impedence of (0 + 5j) W . The bus ‘1’ voltage is 100+30c V, and bus ‘2’ voltage is 100+0c V. The real and reactive power supplied by bus ‘1’ respectively are (A) 1000 W, 268 VAr (B) - 1000 W, - 134 VAr (D) - 276.9 W, 56.7 VAr (C) 276.9 W, - 56.7 VAr

YEAR 2010

TWO MARKS

EE SP 4.41

A three-phase, 33 kV oil circuit breaker is rated 1200 A, 2000 MVA, 3 s. The symmetrical breaking current is (A) 1200 A (B) 3600 A (C) 35 kA (D) 104.8 kA

EE SP 4.42

Consider a stator winding of an alternator with an internal high-resistance ground fault. The currents under the fault condition are as shown in the figure The winding is protected using a differential current scheme with current transformers of ratio 400/5 A as shown. The current through the operating coils is

(A) 0.1875 A (C) 0.375 A

(B) 0.2 A (D) 60 kA


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EE SP 4.44

Power Systems

The zero-sequence circuit of the three phase transformer shown in the figure is

nodia

A 50 Hz synchronous generator is initially connected to a long lossless transmission line which is open circuited at the receiving end. With the field voltage held constant, the generator is disconnected from the transmission line. Which of the following may be said about the steady state terminal voltage and field current of the generator ?

(A) The magnitude not change. (B) The magnitude change. (C) The magnitude increases (D) The magnitude decreases. EE SP 4.45

Chapter 4

of terminal voltage decreases, and the field current does of terminal voltage increases, and the field current does not of terminal voltage increases, and the field current of terminal voltage does not change and the field current

Consider a three-phase, 50 Hz, 11 kV distribution system. Each of the conductors is suspended by an insulator string having two identical porcelain insulators. The self capacitance of the insulator is 5 times the shunt capacitance between the link and the ground, as shown in the figures. The voltages across the two insulators are


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Power Systems

(A) e1 = 3.74 kV, e2 = 2.61 kV (C) e1 = 6.0 kV, e2 = 4.23 kV EE SP 4.46

Page 213

(B) e1 = 3.46 kV, e2 = 2.89 kV (D) e1 = 5.5 kV, e2 = 5.5 kV

Consider a three-core, three-phase, 50 Hz, 11 kV cable whose conductors are denoted as R, Y and B in the figure. The inter-phase capacitance(C1 ) between each line conductor and the sheath is 0.4 mF . The per-phase charging current is

nodia (A) 2.0 A (C) 2.7 A EE SP 4.47

(B) 2.4 A (D) 3.5 A

For the power system shown in the figure below, the specifications of the components are the following : G1 : 25 kV, 100 MVA, X = 9% G2 : 25 kV, 100 MVA, X = 9% T1 : 25 kV/220 kV, 90 MVA, X = 12% T2 : 220 kV/25 kV, 90 MVA, X = 12% Line 1: 200 kV, X = 150 ohms

Choose 25 kV as the base voltage at the generator G1 , and 200 MVA as the MVA base. The impedance diagram is


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Power Systems

Chapter 4

nodia YEAR 2009

ONE MARK

EE SP 4.48

Out of the following plant categories (i) Nuclear (ii) Run-of-river (iii) Pump Storage (iv) Diesel The base load power plant are (A) (i) and (ii) (B) (ii) and (iii) (C) (i), (ii) and (iv) (D) (i), (iii) and (iv)

EE SP 4.49

For a fixed value of complex power flow in a transmission line having a sending


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Power Systems

Page 215

end voltage V , the real loss will be proportional to (A) V (B) V 2 (C) 12 V (D) 1 V YEAR 2009 EE SP 4.50

TWO MARKS

For the Y-bus matrix of a 4-bus system given in per unit, the buses having shunt elements are R- 5 2 2.5 0 V W S S 2 - 10 2.5 4 W YBUS = j S 2.5 2.5 - 9 4 W W S S0 4 4 - 8W X T (A) 3 and 4 (B) 2 and 3

nodia (C) 1 and 2

EE SP 4.51

(D) 1, 2 and 4

Match the items in List-I (To) with the items in the List-II (Use) and select the correct answer using the codes given below the lists. List-I

List-II

a.

improve power factor

1.

shunt reactor

b.

reduce the current ripples

2.

shunt capacitor

c.

increase the power flow in line

3.

series capacitor

4.

series reactor

d. reduce the Ferranti effect (A) a " 2, b " 3, c " 4, d " 1 (B) a " 2, b " 4, c " 3, d " 1 (C) a " 4, b " 3, c " 1, d " 2 (D) a " 4, b " 1, c " 3, d " 2 EE SP 4.52

Match the items in List-I (Type of transmission line) with the items in List-II (Type of distance relay preferred) and select the correct answer using the codes given below the lists. List-I

List-II

a.

Short Line

1.

Ohm Relay

b.

Medium Line

2.

Reactance Relay

c.

Long Line

3.

Mho Relay

(B) a " 3, b " 2, c " 1 (D) a " 1, b " 3, c " 2

(A) a " 2, b " 1, c " 3 (C) a " 1, b " 2, c " 3 EE SP 4.53

Three generators are feeding a load of 100 MW. The details of the generators are Rating (MW)

Efficiency (%)

Regulation (Pu.) ( on 100 MVA base)


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Power Systems

Chapter 4

Generator-1

100

20

0.02

Generator-2

100

30

0.04

Generator-3

100

40

0.03

In the event of increased load power demand, which of the following will happen ? (A) All the generator will share equal power (B) Generator-3 will share more power compared to Generator-1 (C) Generator-1 will share more power compared to Generator-2 (D) Generator-2 will share more power compared to Generator-3 EE SP 4.54

A 500 MW, 21 kV, 50 Hz, 3-phase, 2-pole synchronous generator having a rated p.f = 0.9, has a moment of inertia of 27.5 # 103 kg-m2 .The inertia constant (H ) will be (A) 2.44 s (B) 2.71 s

nodia (C) 4.88 s

(D) 5.42 s

YEAR 2008

EE SP 4.55

ONE MARK

A two machine power system is shown below. The Transmission line XY has positive sequence impedance of Z1 W and zero sequence impedance of Z0 W

An ‘a’ phase to ground fault with zero fault impedance occurs at the centre of the transmission line. Bus voltage at X and line current from X to F for the phase ‘a’, are given by Va Volts and Ia amperes, respectively. Then, the impedance measured by the ground distance relay located at the terminal X of line XY will be given by (B) ^Z0 /2h W (A) ^Z1 /2h W (D) ^Va /Ia h W (C) (Z0 + Z1) /2 W EE SP 4.56

An extra high voltage transmission line of length 300 km can be approximate by a lossless line having propagation constant b = 0.00127 radians per km. Then the percentage ratio of line length to wavelength will be given by (A) 24.24 % (B) 12.12 % (C) 19.05 % (D) 6.06 %

EE SP 4.57

A-3-phase transmission line is shown in figure :


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Power Systems

Page 217

Voltage drop across the transmission line is given by the following equation : VR V V R R S3 Va W SZs Zm Zm WSIa W S3 Vb W = SZm Zs Zm WSIb W SS3 V WW SSZ Z Z WWSSI WW c s m m c X X T T T Shunt capacitance of the X line can be neglect. If the has positive sequence impedance of 15 W and zero sequence impedance of 48 W, then the values of Zs and Zm will be (A) Zs = 31.5 W; Zm = 16.5 W (B) Zs = 26 W; Zm = 11 W

nodia (D) Zs = 11 W; Zm = 26 W

YEAR 2008

EE SP 4.58

TWO MARKS

Voltages phasors at the two terminals of a transmission line of length 70 km have a magnitude of 1.0 per unit but are 180 degree out of phase. Assuming that the maximum load current in the line is 1/5th of minimum 3-phase fault current. Which one of the following transmission line protection schemes will not pick up for this condition ? (A) Distance protection using ohm relay with zoen-1 set to 80% of the line impedance. (B) Directional over current protection set to pick up at 1.25 times the maximum load current (C) Pilot relaying system with directional comparison scheme (D) Pilot relaying system with segregated phase comparison scheme

EE SP 4.59

A loss less transmission line having Surge Impedance Loading (SIL) of 2280 MW is provided with a uniformly distributed series capacitive compensation of 30%. Then, SIL of the compensated transmission line will be (A) 1835 MW (B) 2280 MW (C) 2725 MW (D) 3257 MW

EE SP 4.60

A loss less power system has to serve a load of 250 MW. There are tow generation (G 1 and G 2 ) in the system with cost curves C1 and C2 respectively defined as follows ; 2 C1 (PG1) = PG1 + 0.055 # PG1 2 C2 (PG2) = 3PG2 + 0.03 # PG2 Where PG1 and PG2 are the MW injections from generator G 1 and G 2 respectively. Thus, the minimum cost dispatch will be


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Power Systems

Chapter 4

(A) PG1 = 250 MW; PG2 = 0 MW (B) PG1 = 150 MW; PG2 = 100 MW (C) PG1 = 100 MW; PG2 = 150 MW (D) PG1 = 0 MW; PG2 = 250 MW EE SP 4.61

A loss less single machine infinite bus power system is shown below :

The synchronous generator transfers 1.0 per unit of power to the infinite bus. The critical clearing time of circuit breaker is 0.28 s. If another identical synchronous generator is connected in parallel to the existing generator and each generator is scheduled to supply 0.5 per unit of power, then the critical clearing time of the circuit breaker will (A) reduce to 0.14 s (B) reduce but will be more than 0.14 s

nodia (C) remain constant at 0.28 s (D) increase beyond 0.28 s

EE SP 4.62

Single line diagram of a 4-bus single source distribution system is shown below. Branches e1, e2, e3 and e4 have equal impedances. The load current values indicated in the figure are in per unit.

Distribution company’s policy requires radial system operation with minimum loss. This can be achieved by opening of the branch (B) e2 (A) e1 (C) e3 (D) e4



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Power Systems

Page 219

Given that: Vs1 = Vs2 = 1 + j0 p.u ; The positive sequence impedance are Zs1 = Zs2 = 0.001 + j0.01 p.u and ZL = 0.006 + j0.06 p.u 3-phase Base MVA = 100 voltage base = 400 kV(Line to Line) Nominal system frequency = 50 Hz. The reference voltage for phase ‘a’ is defined as V (t) = Vm cos (wt). A symmetrical three phase fault occurs at centre of the line, i.e. point ‘F’ at time ‘t 0 ’. The positive sequence impedance from source S1 to point ‘F’ equals 0.004 + j0.04 p.u. The wave form corresponding to phase ‘a’ fault current from bus X reveals that decaying d.c. offset current is negative and in magnitude at its maximum initial value, Assume that the negative sequence impedances are equal to positive sequence impedance and the zero sequence impedances are three times positive sequence impedances.

EE SP 4.63

EE SP 4.64

nodia The instant (t0) of the fault will be (A) 4.682 ms

(B) 9.667 ms

(C) 14.667 ms

(D) 19.667 ms

The rms value of the component of fault current (If ) will be (A) 3.59 kA (B) 5.07 kA (C) 7.18 kA (D) 10.15 kA

EE SP 4.65

Instead of the three phase fault, if a single line to ground fault occurs on phase ‘a’ at point ‘F’ with zero fault impedance, then the rms of the ac component of fault current (Ix) for phase ‘a’ will be (A) 4.97 p.u (B) 7.0 p.u (C) 14.93 p.u (D) 29.85 p.u YEAR 2007

EE SP 4.66

ONE MARK

Consider the transformer connections in a part of a power system shown in the figure. The nature of transformer connections and phase shifts are indicated for all but one transformer. Which of the following connections, and the corresponding phase shift q, should be used for the transformer between A and B ?


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Power Systems

Chapter 4

(A) Star-star (q = 0%) (B) Star-Delta (q =- 30%) (C) Delta-star (q = 30%)

nodia (D) Star-Zigzag (q = 30%)

EE SP 4.67

The incremental cost curves in Rs/MWhr for two generators supplying a common load of 700 MW are shown in the figures. The maximum and minimum generation limits are also indicated. The optimum generation schedule is :

(A) Generator A : 400 MW, Generator B : 300 MW (B) Generator A : 350 MW, Generator B : 350 MW (C) Generator A : 450 MW, Generator B : 250 MW (D) Generator A : 425 MW, Generator B : 275 MW EE SP 4.68

Two regional systems, each having several synchronous generators and loads are inter connected by an ac line and a HVDC link as shown in the figure. Which of the following statements is true in the steady state :


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Power Systems

Page 221

(A) Both regions need not have the same frequency (B) The total power flow between the regions (Pac + Pdc) can be changed by controlled the HDVC converters alone (C) The power sharing between the ac line and the HVDC link can be changed by controlling the HDVC converters alone. (D) The directions of power flow in the HVDC link (Pdc ) cannot be reversed EE SP 4.69

Considered a bundled conductor of an overhead line consisting of three identical sub-conductors placed at the corners of an equilateral triangle as shown in the figure. If we neglect the charges on the other phase conductor and ground, and assume that spacing between sub-conductors is much larger than their radius, the maximum electric field intensity is experienced at

nodia (A) Point X (C) Point Z

(B) Point Y (D) Point W


TWO MARKS

The figure below shows a three phase self-commutated voltage source converter connected to a power system. The converter’s dc bus capacitor is marked as C in the figure. The circuit in initially operating in steady state with d = 0 and the capacitor dc voltage is equal to Vdc0 . You may neglect all losses and harmonics. What action should be taken to increase the capacitor dc voltage slowly to a new steady state value.

(A) Make d positive and maintain it at a positive value (B) Make d positive and return it to its original value (C) Make d negative and maintain it at a negative value (D) Make d negative and return it to its original value EE SP 4.71

The total reactance and total suspectance of a lossless overhead EHV line, operating at 50 Hz, are given by 0.045 pu and 1.2 pu respectively. If the velocity of wave propagation is 3 # 105 km/s, then the approximate length of the line is (B) 172 km (A) 122 km (C) 222 km (D) 272 km


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EE SP 4.73

Power Systems

Consider the protection system shown in the figure below. The circuit breakers numbered from 1 to 7 are of identical type. A single line to ground fault with zero fault impedance occurs at the midpoint of the line (at point F), but circuit breaker 4 fails to operate (‘‘Stuck breaker’’). If the relays are coordinated correctly, a valid sequence of circuit breaker operation is

nodia (A) 1, 2, 6, 7, 3, 5

(B) 1, 2, 5, 5, 7, 3

(C) 5, 6, 7, 3, 1, 2

(D) 5, 1, 2, 3, 6, 7

A three phase balanced star connected voltage source with frequency w rad/s is connected to a star connected balanced load which is purely inductive. The instantaneous line currents and phase to neutral voltages are denoted by (ia, ib, ic) and (Van, Vbn, Vcn) respectively, and their rms values are denoted by V and I . R V 1 - 13 W RSia VW S 0 3 1 WS W If R = 8Van Vbn Vcn B S- 13 0 i , then the magnitude of 3 W b S 1 SSi WW 1 0 W c S 3 - 3 T XT X of R is (A) 3VI (C) 0.7VI

EE SP 4.74

Chapter 4

(B) VI (D) 0

Consider a synchronous generator connected to an infinite bus by two identical parallel transmission line. The transient reactance ‘x’ of the generator is 0.1 pu and the mechanical power input to it is constant at 1.0 pu. Due to some previous disturbance, the rotor angle (d) is undergoing an undamped oscillation, with the maximum value of d (t) equal to 130% .One of the parallel lines trip due to the relay maloperation at an instant when d (t) = 130% as shown in the figure. The maximum value of the per unit line reactance, x such that the system does not lose synchronism subsequent to this tripping is

(A) 0.87 (C) 0.67

(B) 0.74 (D) 0.54


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EE SP 4.76

Power Systems

Page 223

Suppose we define a sequence transformation between ‘‘a-b-c’’ and ‘‘p-n-o’’ variables as follows : R V VR V R Sfa W S 1 1 1W Sfp W 2p Sfb W = k Sa2 a 1W Sfn W where a = e j 3 and k and is a constant SSf WW SSa a2 1WW SSf WW VR V R V R V R R V o c Va W Vp W S0.5 0 0 W Sip W Sia W S S T X XT X T Now, if it is given that : SVn W = S 0 0.5 0 W Sin W and SVb W = Z Sib W then, SSi WW SSV WW SS 0 0 2.0 WW SSi WW SSV WW 0 c o c XT X R T X TV X VT X T R S1.0 0.5 0.5 W S 1.0 0.5 0.75 W (B) Z = S0.5 1.0 0.5 W (A) Z = S0.75 1.0 0.5 W SS0.5 0.5 1.0 WW SS 0.5 0.75 1.0 WW X V X T R T R V . . . . 5 - 0. 5 W 1 0 . 1 0 0 75 0 0 5 W S 2S (C) Z = 3k2 S 0.5 1.0 0.75 W (D) Z = k S- 0.5 1.0 - 0.5 W 3S SS0.75 0.5 1.0 WW S- 0.5 - 0.5 1.0 WW X X T T Consider the two power systems shown in figure A below, which are initially not interconnected, and are operating in steady state at the same frequency. Separate load flow solutions are computed individually of the two systems, corresponding to this scenario. The bus voltage phasors so obtain are indicated on figure A. These two isolated systems are now interconnected by a short transmission line as shown in figure B, and it is found that P1 = P2 = Q1 = Q2 = 0 .

nodia

The bus voltage phase angular difference between generator bus X and generator bus Y after interconnection is (A) 10c (B) 25c


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Power Systems

(C) - 30c EE SP 4.77

Chapter 4

(D) 30c

A 230 V (Phase), 50 Hz, three-phase, 4-wire, system has a phase sequence ABC. A unity power-factor load of 4 kW is connected between phase A and neutral N. It is desired to achieve zero neutral current through the use of a pure inductor and a pure capacitor in the other two phases. The value of inductor and capacitor is (A) 72.95 mH in phase C and 139.02 mF in Phase B (B) 72.95 mH in Phase B and 139.02 mF in Phase C (C) 42.12 mH in Phase C and 240.79 mF in Phase B (D) 42.12 mH in Phase B and 240.79 mF in Phase C

EE SP 4.78

An isolated 50 Hz synchronous generator is rated at 15 MW which is also the maximum continuous power limit of its prime mover. It is equipped with a speed governor with 5% droop. Initially, the generator is feeding three loads of 4 MW each at 50 Hz. One of these loads is programmed to trip permanently if the frequency falls below 48 Hz .If an additional load of 3.5 MW is connected then the frequency will settle down to (A) 49.417 Hz (B) 49.917 Hz (C) 50.083 Hz (D) 50.583 Hz

nodia YEAR 2006

EE SP 4.79

ONE MARK

The concept of an electrically short, medium and long line is primarily based on the (A) nominal voltage of the line (B) physical length of the line (C) wavelength of the line

(D) power transmitted over the line

EE SP 4.80

Keeping in view the cost and overall effectiveness, the following circuit breaker is best suited for capacitor bank switching (A) vacuum (B) air blast (C) SF6 (D) oil

EE SP 4.81

In a biased differential relay the bias is defined as a ratio of (A) number of turns of restraining and operating coil (B) operating coil current and restraining coil current (C) fault current and operating coil current (D) fault current and restraining coil current

EE SP 4.82

An HVDC link consist of rectifier, inverter transmission line and other equipments. Which one of the following is true for this link ? (A) The transmission line produces/ supplies reactive power (B) The rectifier consumes reactive power and the inverter supplies reactive power from/ to the respective connected AC systems (C) Rectifier supplies reactive power and the inverted consumers reactive power to/ from the respective connected AC systems


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Power Systems

Page 225

(D) Both the converters (rectifier and inverter) consume reactive power from the respective connected AC systems YEAR 2006

TWO MARKS

EE SP 4.83

The A, B , C , D constants of a 220 kV line are : A = D = 0.94+1c, B = 130+73c , C = 0.001+90c If the sending end voltage of the line for a given load delivered at nominal voltage is 240 kV, the % voltage regulation of the line is (A) 5 (B) 9 (C) 16 (D) 21

EE SP 4.84

A single phase transmission line and a telephone line are both symmetrically strung one below the other, in horizontal configurations, on a common tower, The shortest and longest distances between the phase and telephone conductors are 2.5 m and 3 m respectively. The voltage (volt/km) induced in the telephone circuit, due to 50 Hz current of 100 amps in the power circuit is (A) 4.81 (B) 3.56 (C) 2.29 (D) 1.27

EE SP 4.85

nodia

Three identical star connected resistors of 1.0 pu are connected to an unbalanced 3-phase supply. The load neutral is isolated. The symmetrical components of the line voltages in pu. are: Vab = X+q1 , Vab = Y+q2 . If all the pu calculations are with the respective base values, the phase to neutral sequence voltages are (A) Van = X+ (q1 + 30c),Van = Y (q2 - 30c) (B) Van = X+ (q1 - 30c),Van = Y+ (q2 + 30c) (C) Van = 1 X+ (q1 - 30c),Van = 1 Y+ (q2 - 30c) 3 3 1 1 (D) Van = Y+ (q2 - 60c) X+ (q1 - 60c),Van = 3 3 1

1

2

1

2

2

1

2

1

2

EE SP 4.86

A generator is connected through a 20 MVA, 13.8/138 kV step down transformer, to a transmission line. At the receiving end of the line a load is supplied through a step down transformer of 10 MVA, 138/69 kV rating. A 0.72 pu. load, evaluated on load side transformer ratings as base values , is supplied from the above system. For system base values of 10 MVA and 69 kV in load circuit, the value of the load (in per unit) in generator will be (A) 36 (B) 1.44 (C) 0.72 (D) 0.18

EE SP 4.87

The Gauss Seidel load flow method has following disadvantages. Tick the incorrect statement. (A) Unreliable convergence (B) Slow convergence (C) Choice of slack bus affects convergence (D) A good initial guess for voltages is essential for convergence


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Power Systems

Chapter 4

Common Data For Q. 88 and 89 A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30c. The system frequency is 50 Hz. EE SP 4.88

EE SP 4.89

The initial accelerating power (in pu) will be (A) 1.0 (B) 0.6 (C) 0.56 (D) 0.4

nodia

If the initial accelerating power is X pu, the initial acceleration in elect-deg/sec, and the inertia constant in MJ-sec/elect-deg respectively will be (B) 1800X , 0.056 (A) 31.4X , 18 (C) X/1800, 0.056 (D) X/31.4, 18


For a power system the admittance and impedance matrices for the fault studies are as follows. V V R R S- j8.75 j1.25 j2.50 W Sj0.16 j0.08 j0.12 W Z bus = Sj0.08 j0.24 j0.16 W Ybus = S j1.25 - j6.25 j2.50 W SS j2.50 - j2.50 - j5.00 WW SSj0.12 j0.16 j0.34 WW X X T T The pre-fault voltages are 1.0 pu. at all the buses. The system was unloaded prior to the fault. A solid 3-phase fault takes place at bus 2.

EE SP 4.90

The post fault voltages at buses 1 and 3 in per unit respectively are (A) 0.24, 0.63 (B) 0.31, 0.76 (C) 0.33, 0.67 (D) 0.67, 0.33

EE SP 4.91

The per unit fault feeds from generators connected to buses 1 and 2 respectively are (A) 1.20, 2.51 (B) 1.55, 2.61 (C) 1.66, 2.50 (D) 5.00, 2.50

EE SP 4.92

A 400 V, 50 Hz, three phase balanced source supplies power to a star connected load whose rating is 12 3 kVA, 0.8 pf (lag). The rating (in kVAR) of the delta connected (capacitive) reactive power bank necessary to bring the pf to unity is (A) 28. 78 (B) 21.60 (C) 16.60 (D) 12.47 YEAR 2005

EE SP 4.93

ONE MARK

The p.u. parameter for a 500 MVA machine on its own base are: inertia, M = 20 p.u. ; reactance, X = 2 p.u.


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Power Systems

Page 227

The p.u. values of inertia and reactance on 100 MVA common base, respectively, are (A) 4, 0.4 (B) 100, 10 (C) 4, 10 (D) 100, 0.4 EE SP 4.94

An 800 kV transmission line has a maximum power transfer capacity of P . If it is operated at 400 kV with the series reactance unchanged, the new maximum power transfer capacity is approximately (A) P (B) 2P (C) P/2 (D) P/4

EE SP 4.95

The insulation strength of an EHV transmission line is mainly governed by (A) load power factor (B) switching over-voltages (C) harmonics (D) corona

EE SP 4.96

nodia High Voltage DC (HVDC) transmission is mainly used for (A) bulk power transmission over very long distances (C) inter-connecting two systems with same nominal frequency (C) eliminating reactive power requirement in the operation (D) minimizing harmonics at the converter stations YEAR 2005

TWO MARKS

EE SP 4.97

The parameters of a transposed overhead transmission line are given as : Self reactance XS = 0.4W/km and Mutual reactance Xm = 0.1W/km The positive sequence reactance X1 and zero sequence reactance X0 , respectively in W/km are (A) 0.3, 0.2 (B) 0.5, 0.2 (C) 0.5, 0.6 (D) 0.3, 0.6

EE SP 4.98

At an industrial sub-station with a 4 MW load, a capacitor of 2 MVAR is installed to maintain the load power factor at 0.97 lagging. If the capacitor goes out of service, the load power factor becomes (A) 0.85 (B) 1.00 (C) 0.80 lag (D) 0.90 lag

EE SP 4.99

The network shown in the given figure has impedances in p.u. as indicated. The diagonal element Y22 of the bus admittance matrix YBUS of the network is

(A) - j19.8 (C) + j0.2 EE SP 4.100

(B) + j20.0 (D) - j19.95

A load centre is at an equidistant from the two thermal generating stations G1


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Power Systems

Chapter 4

and G2 as shown in the figure. The fuel cost characteristic of the generating stations are given by F1 = a + bP1 + cP12 Rs/hour F2 = a + bP2 + 2cP22 Rs/ hour

Where P1 and P2 are the generation in MW of G1 and G2 , respectively. For most economic generation to meet 300 MW of load P1 and P2 respectively, are (A) 150, 150 (B) 100, 200 (C) 200, 100 (D) 175, 125

EE SP 4.101

nodia

Two networks are connected in cascade as shown in the figure. With usual notations the equivalent A, B, C and D constants are obtained. Given that, C = 0.025+45c, the value of Z2 is

(A) 10+30c W (C) 1 W EE SP 4.102

(B) 40+ - 45c W (D) 0 W

A generator with constant 1.0 p.u. terminal voltage supplies power through a step-up transformer of 0.12 p.u. reactance and a double-circuit line to an infinite bus bar as shown in the figure. The infinite bus voltage is maintained at 1.0 p.u. Neglecting the resistances and suspectances of the system, the steady state stability power limit of the system is 6.25 p.u. If one of the double-circuit is tripped, then resulting steady state stability power limit in p.u. will be

(A) 12.5 p.u. (C) 10.0 p.u.

(B) 3.125 p.u. (D) 5.0 p.u.

Common Data For Q. 103 and 104 At a 220 kV substation of a power system, it is given that the three-phase fault level is 4000 MVA and single-line to ground fault level is 5000 MVA Neglecting the resistance and the shunt suspectances of the system. EE SP 4.103

The positive sequence driving point reactance at the bus is (A) 2.5 W (B) 4.033 W


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Power Systems

(C) 5.5 W EE SP 4.104

Page 229

(D) 12.1 W

The zero sequence driving point reactance at the bus is (A) 2.2 W (B) 4.84 W (C) 18.18 W (D) 22.72 W YEAR 2004

EE SP 4.105

ONE MARK

Total instantaneous power supplied by a 3-phase ac supply to a balanced R-L load is (A) zero (B) constant

nodia (C) pulsating with zero average

(D) pulsating with the non-zero average

EE SP 4.106

The rated voltage of a 3-phase power system is given as (A) rms phase voltage (B) peak phase voltage (C) rms line to line voltage

(D) peak line to line voltage

EE SP 4.107

The phase sequences of the 3-phase system shown in figure is

(A) RYB (B) RBY (C) BRY (D) YBR EE SP 4.108

In the thermal power plants, the pressure in the working fluid cycle is developed by (A) condenser (B) super heater (C) feed water pump (D) turbine

EE SP 4.109

For harnessing low variable waterheads, the suitable hydraulic turbine with high percentage of reaction and runner adjustable vanes is (A) Kaplan (B) Francis (C) Pelton (D) Impeller


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Power Systems

Chapter 4

The transmission line distance protection relay having the property of being inherently directional is (A) impedance relay (B) MHO relay (C) OHM relay (D) reactance relay YEAR 2004

EE SP 4.111

A 800 kV transmission line is having per phase line inductance of 1.1 mH/km and per phase line capacitance of 11.68 nF/km. Ignoring the length of the line, its ideal power transfer capability in MW is (A) 1204 MW (B) 1504 MW

nodia (C) 2085 MW

EE SP 4.112

EE SP 4.113

TWO MARKS

(D) 2606 MW

A 110 kV, single core coaxial, XLPE insulated power cable delivering power at 50 Hz, has a capacitance of 125 nF/km. If the dielectric loss tangent of XLPE is 2 # 10 - 4 , then dielectric power loss in this cable in W/km is (A) 5.0 (B) 31.7 (C) 37.8 (D) 189.0

A lightning stroke discharges impulse current of 10 kA (peak) on a 400 kV transmission line having surge impedance of 250 W The magnitude of transient over-voltage travelling waves in either direction assuming equal distribution from the point of lightning strike will be (A) 1250 kV (B) 1650 kV (C) 2500 kV (D) 2900 kV

EE SP 4.114

The generalized circuit constants of a 3-phase, 220 kV rated voltage, medium length transmission line are A = D = 0.936 + j0.016 = 0.936+0.98c B = 35.5 + j138 = 142.0+76.4c W C = (- 5.18 + j914) # 10 - 6 W If the load at the receiving end is 50 MW at 220 kV with a power factor of 0.9 lagging, then magnitude of line to line sending end voltage should be (A) 133.23 kV (B) 220.00 kV (C) 230. 78 kV (D) 246.30 kV

EE SP 4.115

pu. [equivalent to A new generator having Eg = 1.4+30c (1.212 + j0.70) pu] and synchronous reactance 'XS ' of 1.0 pu on the system base, is to be connected to a bus having voltage Vt , in the existing power system. This existing power system can be represented by Thevenin’s voltage Eth = 0.9+0c pu in series with Thevenin’s impedance Zth = 0.25+90c pu. The magnitude of the bus voltage Vt of the system in pu will be (A) 0.990 (B) 0.973 (C) 0.963 (D) 0.900


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Power Systems

A 3-phase generator rated at 110 MVA, 11 kV is connected through circuit breakers to a transformer . The generator is having direct axis sub-transient reactance X''d = 19% , transient reactance X'd = 26% and synchronous reactance =130%. The generator is operating at no load and rated voltage when a three phase short circuit fault occurs between the breakers and the transformer . The magnitude of initial symmetrical rms current in the breakers will be (A) 4.44 kA (B) 22.20 kA (C) 30.39 kA

EE SP 4.117

Page 231

(D) 38.45 kA

A 3-phase transmission line supplies 3-connected load Z . The conductor ‘c’ of the line develops an open circuit fault as shown in figure. The currents in the lines are as shown on the diagram. The +ve sequence current component in line ‘a’ will be

nodia (A) 5.78+ - 30c (C) 6.33+90c

(B) 5.78+90c (D) 10.00+ - 30c

EE SP 4.118

A 500 MVA, 50 Hz, 3-phase turbo-generator produces power at 22 kV. Generator is Y-connected and its neutral is solidly grounded. It sequence reactances are X1 = X2 = 0.15 pu and X 0 = 0.05 pu.It is operating at rated voltage and disconnected from the rest of the system (no load). The magnitude of the sub-transient line current for single line to ground fault at the generator terminal in pu will be (A) 2.851 (B) 3.333 (C) 6.667 (D) 8.553

EE SP 4.119

A 50 Hz, 4-pole, 500 MVA, 22 kV turbo-generator is delivering rated megavoltamperes at 0.8 power factor. Suddenly a fault occurs reducing in electric power output by 40%. Neglect losses and assume constant power input to the shaft. The accelerating torque in the generator in MNm at the time of fault will be (A) 1.528 (B) 1.018 (C) 0.848 (D) 0.509

EE SP 4.120

A hydraulic turbine having rated speed of 250 rpm is connected to a synchronous generator. In order to produce power at 50 Hz, the number of poles required in the generator are (A) 6 (B) 12 (C) 16 (D) 24


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Power Systems

YEAR 2003 EE SP 4.121

Chapter 4

ONE MARK

Bundled conductors are mainly used in high voltage overhead transmission lines to (A) reduces transmission line losses (B) increase mechanical strength of the line (C) reduce corona (D) reduce sag

EE SP 4.122

EE SP 4.123

A power system consist of 300 buses out of which 20 buses are generator bus, 25 buses are the ones with reactive power support and 15 buses are the ones with fixed shunt capacitors. All the other buses are load buses. It is proposed to perform a load flow analysis in the system using Newton-Raphson method. The size of the Newton Raphson Jacobian matrix is (A) 553 # 553 (B) 540 # 540 (C) 555 # 555 (D) 554 # 554

nodia

Choose two appropriate auxiliary components of a HVDC transmission system from the following P. D.C line inductor Q. A.C line inductor R. Reactive power sources

S. Distance relays on D.C line T. Series capacitance on A.C. line (A) P and Q (B) P and R (C) Q and S (D) S and T

EE SP 4.124

A round rotor generator with internal voltage E1 = 2.0 pu and X = 1.1 pu is connected to a round rotor synchronous motor with internal voltage E2 = 1.3 pu and X = 1.2 pu. The reactance of the line connecting the generator to the motor is 0.5 pu. When the generator supplies 0.5 pu power, the rotor angle difference between the machines will be (A) 57.42c (B) 1c (C) 32.58c (D) 122.58c

EE SP 4.125

The interrupting time of a circuit breaker is the period between the instant of (A) initiation of short circuit and the arc extinction on an opening operation (B) energizing of the trip circuit and the arc extinction on an opening operation (C) initiation of short circuit and the parting of primary arc contacts (D) energizing of the trip circuit and the parting of primary arc contacts


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Power Systems

YEAR 2003

Page 233

TWO MARKS

EE SP 4.126

The ABCD parameters of a 3-phase overhead transmission line are A = D = 0.9+0c, B = 200+90c W and C = 0.95 # 10 - 3 +90% S . At no-load condition a shunt inductive, reactor is connected at the receiving end of the line to limit the receiving-end voltage to be equal to the sending-end voltage. The ohmic value of the reactor is (A) 3 W (B) 2000 W (C) 105.26 W (D) 1052.6 W

EE SP 4.127

A surge of 20 kV magnitude travels along a lossless cable towards its junction with two identical lossless overhead transmission lines. The inductance and the capacitance of the cable are 0.4 mH and 0.5 mF per km. The inductance and capacitance of the overhead transmission lines are 1.5 mH and 0.015 mF per km. The magnitude of the voltage at the junction due to surge is (A) 36.72 kV (B) 18.36 kV (C) 6.07 kV (D) 33.93 kV

EE SP 4.128

nodia

A dc distribution system is shown in figure with load current as marked. The two ends of the feeder are fed by voltage sources such that VP - VQ = 3 V. The value of the voltage VP for a minimum voltage of 220 V at any point along the feeder is

(A) 225.89 V (B) 222.89 V (C) 220.0 V (D) 228.58 V EE SP 4.129

A balanced delta connected load of (8 + j6) W per phase is connected to a 400 V, 50 Hz, 3-phase supply lines. If the input power factor is to be improved to 0.9 by connecting a bank of star connected capacitor the required kVAR of the of the bank is (A) 42.7 (B) 10.2 (C) 28.8 (D) 38.4

EE SP 4.130

A 3-phase 11 kV generator feeds power to a constant power unity power factor load of 100 MW through a 3-phase transmission line. The line-to line voltage at the terminals of the machine is maintained constant at 11 kV. The per unit positive sequence impedance of the line based on 100 MVA and 11 kV is j0.2 . The line to line voltage at the load terminals is measured to be less than 11 kV. The total reactive power to be injected at the terminals of the load to increase


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Power Systems

Chapter 4

the line-to-line voltage at the load terminals to 11 kV is (A) 100 MVAR (B) 10.1 MVAR (C) - 100 MVAR (D) - 10.1 MVAR EE SP 4.131

EE SP 4.132

The bus impedance matrix of a 4-bus power system is given by Rj0.3435 j0.2860 j0.2723 j0.2277 V W S Sj0.2860 j0.3408 j0.2586 j0.2414 W Z bus = S j0.2723 j0.2586 j0.2791 j0.2209 W W S Sj0.2277 j0.2414 j0.2209 j0.2791 W X T A branch having an impedance of j0.2 W is connected between bus 2 and the reference. Then the values of Z22,new and Z23,new of the bus impedance matrix of the modified network are respectively (A) j0.5408 W and j0.4586 W (B) j0.1260 W and j0.0956 W (C) j0.5408 W and j0.0956 W (D) j0.1260 W and j0.1630 W

nodia

A 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line. The per unit positive-sequence, negative-sequence and zero-sequence impedances of the alternator are j0.1, j0.1 and j0.04 respectively. The neutral of the alternator is connected to ground through an inductive reactor of j0.05 p.u. The per unit positive-, negative- and zero-sequence impedances of transmission line are j0.1, j0.1 and j0.3, respectively. All per unit values are based on the machine ratings. A solid ground fault occurs at one phase of the far end of the transmission line. The voltage of the alternator neutral with respect to ground during the fault is (A) 513.8 V (B) 889.9 V (C) 1112.0 V (D) 642.2 V

EE SP 4.133

Incremental fuel costs (in some appropriate unit) for a power plant consisting of three generating units are IC1 = 20 + 0.3P1, IC2 = 30 + 0.4P2, IC3 = 30 Where P1 is the power in MW generated by unit i for i = 1, 2 and 3. Assume that all the three units are operating all the time. Minimum and maximum loads on each unit are 50 MW and 300 MW respectively. If the plant is operating on economic load dispatch to supply the total power demand of 700 MW, the power generated by each unit is (A) P1 = 242.86 MW; P2 = 157.14 MW; and P3 = 300 MW (B) P1 = 157.14 MW; P2 = 242.86 MW; and P3 = 300 MW (C) P1 = 300 MW; P2 = 300 MW; and P3 = 100 MW (D) P1 = 233.3 MW; P2 = 233.3 MW; and P3 = 233.4 MW

EE SP 4.134

A list of relays and the power system components protected by the relays are given in List-I and List-II respectively. Choose the correct match from the four choices given below: List-I P.

List-II

Distance relay

1.

Transformers


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Power Systems

Page 235

Q.

Under frequency relay

2.

Turbines

R.

Differential relay

3.

Busbars

S.

Buchholz relay

4.

Shunt capacitors

5.

Alternators

6.

Transmission lines

Codes: (A) (B) (C) (D) EE SP 4.135

EE SP 4.136

P 6 4 5 6

Q 5 3 2 4

R 3 2 1 5

S 1 1 6 3

nodia

A generator delivers power of 1.0 p.u. to an infinite bus through a purely reactive network. The maximum power that could be delivered by the generator is 2.0 p.u. A three-phase fault occurs at the terminals of the generator which reduces the generator output to zero. The fault is cleared after tc second. The original network is then restored. The maximum swing of the rotor angle is found to be dmax = 110 electrical degree. Then the rotor angle in electrical degrees at t = tc is (A) 55 (B) 70 (C) 69.14 (D) 72.4 A three-phase alternator generating unbalanced voltages is connected to an unbalanced load through a 3-phase transmission line as shown in figure. The neutral of the alternator and the star point of the load are solidly grounded. The phase voltages of the alternator are Ea = 10+0c V, Eb = 10+ - 90c V, Ec = 10+120c V . The positive-sequence component of the load current is

(A) 1.310+ - 107c A (C) 0.996+ - 120c A

(B) 0.332+ - 120c A (D) 3.510+ - 81c A ***********


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Power Systems

Chapter 4

SOLUTION SOL 4.1

Correct option ( C). We have the given system as

nodia

As fault always require/demand reactive power, so voltage should lead current. Using this condition, we have the following observations for the given system: At fault F2 which is not near to source require reactive power, hence VF2 leads IF2 Fault F1 is near the source, so the current will lead voltage for active requirements, i.e. VF1 lags IF1

SOL 4.2

Correct option is (B). As the winding and neutral has affect then we represent transformer in zero sequence network using switch diagram.

Switches 1, 1l " Primary Switches 2, 2l, " Secondary winding 1, 2 " Series switches 1l, 2l " Shunt switches Series switches are closed when transformer is star-connected. Shunt switches are closed when transformer is delta-connected. For ground effect 3X 0 .

Here, on T1 side series switch are closed hence star-connection and grounded due


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Power Systems

Page 237

to 3XGn . On T2 side, shunt switches are closed hence one delta and one star with ground.

SOL 4.3

T1

:

T2

:

Correct option is (C). For the unbalanced three phase system, we have

nodia Ia1 = 1 - 90c pu ; Ib2 = 4 - 150c pu ;

Ic0 = 3 90c pu The phase current is given by Ia = Ia1 + Ia2 + Ia0

...(i)

Now, we have

Ib2 = aIa2 So, for the given system

4 - 150c = 1 120c Ia2

or Also, we have

Ia2 = 4 - 270c

Ia0 = Ib0 = Ic0 = 3 90c Substituting these values in equation (i), we get or So,

Ia = 1 - 90c = Ia1 + 4 - 270c + 3 90c Ia1 = 1 - 90c - 4 - 270c - 3 90c = 8 - 90c Ib1 = a2 Ia1 = 1 240c # 8 - 90c = 8 150c

Thus, we obtain Ib = Ib1 + Ib2 + Ib0 = 8 150c + 4 - 150c + 3 90c = 11.53 154.3c pu SOL 4.4

Correct option is (A). Given the three phase synchronous generator,


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Power Systems

Chapter 4

If all the bulbs glow dark simultaneously, then all the gear and infinite bus are in same phase sequence. But, in the given problem, they are coming in sequence (not simultaneously) La then Lb , and then Lc . So, phase sequence of the generator is opposite to infinite bus system and frequency is high (as dark period is for more time).

SOL 4.5

nodia Correct option is (D). We have the distribution feeder as

Let current I be supplied by source ^s1h, and r be the resistance per unit length. Applying KVL, we get 400 - ^r # 400h I - ^200r hÎ - 200h - ^200r hÎ - 300h - ^200r hÎ - 500h = 400 or 400Ir + 200Ir - 40000r + 200Ir - 60000r + 200Ir - 100000r = 0 or 1000Ir = 200000r So, I = 200 A So, we redraw the current in distribution system as

Thus, the contribution of source ^s1h in 100 A load at P is 0 A , and the contribution of s2 in 100 A load at P is 100 A. SOL 4.6

Correct option is (B). Given the two bus power system,


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Power Systems

Page 239

Now, the current through impedance Z is given by I = V1 - V2 Z For the given system, we have V1 = V1 0c ; V2 = 1 d 2 , z = 0.1 90c So, we obtain the current V 0c - 1 d 2 = 1 + j0.5 I q = 1 0.1 90c V1 0c - 1 d 2 = 0.11 116.56 or V1 - ^cos d 2 - j sin d 2h = 0.11 6cos 116.56c + j sin 116.36c@ Comparing the real and imaginary terms, we get or

- sin d 2 = 0.11 # sin 116.36c - sin d 2 = 0.098

nodia or

d 2 =- 5.65c V1 - cos d 2 = 0.11 cos 116.56c

or Also,

V1 = 1.05

or

SOL 4.7

Correct option is (D). The fuel cost functions for plant P1 and P2 are C1 = 0.05Pg + APg + B C2 = 0.10Pg + 3APg + 2B Given Pg + Pg = 1000 MW dC1 = dC2 = 100 and dPg dPg Also, from equations (i) and (ii), we have dC1 = 2 0.05P + A # g dPg dC2 = 2 0.10P + 3A # g dPg Substituting equation (iii) in above expressions, we get 0.1Pg + A = 100 0.2Pg + 3A = 100 0.3Pg - 0.2Pg = 200 Hence, solving the equations (iii) and (v), we obtain 2 1

1

2 2

1

2

2

1

...(i) ...(ii) ...(iii) ...(iv)

2

1

1

2

2

1

2

1

2

...(v)

Pg = 800 MW and Pg = 200 MW Therefore, the ratio of load shared by plants P1 and P2 is Pg = 800 = 4 200 1 Pg 1

2

1

2

SOL 4.8

Correct answer is 0.226. We have the circuit,


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Power Systems

Chapter 4

The operating time of relays is given as 0.14 # Time multiplier setting (TMS) top = ^plug setting multiplierh0.02 - 1 For relay RB , we have TMS = 0.1

...(i)

PSM = maximum fault current CT ratio 000 2 = = 20 500/5 Substituting the values in equation (i), we get 0.1 = 0.14 # 0.1 top = 0.140# 0.617 ^20h .02 - 1 = 0.226

SOL 4.9

nodia Correct answer is 118.80 .

p.u. value = Acual value in same units Base value in same units Base voltage = Base MVA Base Current 6

Base voltage = 100 # 10 = 228.63 kV 437.38 A 0.9 p.u. = Actual value 228.63 kV Vactual = 228.63 # 0.9 kV = 205.77 kV As load is star connected VL - L = Vacutal = 205.77 kV 3 3 VL - L = 118.80 kV SOL 4.10

Correct option is (C). Shunt reactors are used in high voltage transmission system to limit over voltages at the load side under lightly loaded condition.

SOL 4.11

Correct answer is 0.1092 . As active power in synchronous generator is given by E V d = Load angle P = f sin d Xs 0.6 = 1.3 # 1 sin d { E f = 1.3 pu , V = 1 pu , Xs = 1.1 pu , P = 0.6 pu } 1.1 sin d = 0.6 # 1.1 1.3 # 1 sin d = 0.5077


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Power Systems

Page 241

d = 30.504c Reactive power is given by 2 E V Q = f cos d - V Xs Xs 2 = 1.3 # 1 cos ^30.504ch - 1 1. 1 1. 1

Q = 1.1818 # 0.8616 - 0.9090 Q = 0.1092 pu SOL 4.12

Correct answer is 50. Since the generators are in parallel, they will operate at the same frequency ^ f h at steady loss. Let load on generator G1 is P1 and on G2 is P2 . Then we can write for generator G1 51.5 - f = droop constant P1 51.5 - f =1 P1

nodia

...(i) 51.5 - f = P1 Now in steady state total load is 2.5 MW , so load by second generator is ^2.5 - P1h . For G2 we write 51 - f =1 2.5 - P1

51 - f = 2.5 - P1 Adding eq (i) and (ii), we get

...(ii)

51.5 - f + 51 - f = P1 + 2.5 - P1 102.5 - 2f = 2.5 2f = 100 f = 50 Hz SOL 4.13

Correct option is (B). C = 2pe 0 ln b d2 l d1 d2 = distance b/w two sphere, d1 = radius of first sphere -12 = 2 # p # 8.85 # 10 ln b 600 l 0.8 -12 C = 8.39 # 10 F/m C = 8.39 # 10-12 # 10-3 F Ceq . 8.4 # 10-9 F

SOL 4.14

Correct answer is 15403.698 . For single line-to-ground fault, the symmetrical component of current (in pu) Ea1 1 = = 2.22 pu Ia1 = z 0 + z1 + z2 0.2 + 0.2 + 0.05 The magnitude of fault current I f = 3Ia1 = 3 # 2.22 = 6.67 pu Base value of current


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Power Systems

Chapter 4

100 # 106 = 2309.40 A 3 # 25 # 103 So actual value of fault current Ibase =

I f = 6.67 # 2309.40 = 15403.698 A SOL 4.15

Correct option is (A). For a transmission line to be distortionless, the required condition is R =G L C G = shunt conductance (in siemens/meter) C = capacitance (in Farad/meter) R = Resistance (in ohm/meter)

where

nodia

L = inductance (in henries per meter) So, for the given parameters of long transmission line, the required condition for distortionless transmission becomes r =g c l rc = lg

or

SOL 4.16

SOL 4.17

Correct option is (B). A fully transposed transmission line is completely symmetrically and therefore the phase impedance offered by it is independent of phase sequence of a balanced set of current. In other word impedance offered by it to positive and negative sequence currents are equal. Correct answer is 332. Given, number of PQ buses = 150 number of PV buses = 32 For PQ buses, 2 equations are formed (one for P and other for Q) For PV buses only 1 equation is formed (for Q only) Thus, the minimum number of simultaneous equations, required to obtain the load, is Total number of equations = Number of equation for PQ buses + Number of equation for PV buses. = 2 # (number of PQ buses) + 1 # (number of PV buses) = 2 # (150) + 1 # (32) = 332

SOL 4.18

Correct answer is 187

SOL 4.19

Correct answer is 0.75kVAR. Given the constant voltage load, P1 + jQ1 where 1 kW < P1 < 1.5 kW and 0.5 kVAR # Q1 # 1 kVAR Now, the shunt capacitor is chosen such that reactive power across capacitor


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Power Systems

Page 243

load is Q # 0.25 kVAR Since, we have Q1 = 1 kVAR , therefore, capacitor will supply QC = .75 kVAR i.e. capacitor must supply .75 kVAR . SOL 4.20

Correct answer is 0.23 .

SOL 4.21


SOL 4.22

Correct option is (C). Consider the voltage phase angles at buses 2 and 3 be Q2 and Q 3 since, all the three buses have the equal voltage magnitude which is 1 Pu so, it is a D.C. load flow. The injections at Bus 2 and 3 are respectively P2 = 0.1 pu P3 =- 0.2 pu Therefore, the phase angles are obtained as q2 P2 >q H = 6B@-1 >P H 3 3 where 6B@ is obtained as R V 1 S 1 + 1 W X12 X23 X23 W S 6B@ = S - 1 1 + 1 W S X23 X23 X13 W T X 1 + 1 -1 2 -1 H=> H => -1 1 + 1 -1 2 Its inverse is obtained as -1 2 1 0.1 2 -1 -1 > H H> H = 1> 6B@ = 1 - 2 3 + 1 2 - 0.2 Therefore, q2 0 1 2 1 0.1 1 0 >q H = 3 >+ 1 2H>- 0.2H = 3 >- 0.3H = >- 0.1H 3

nodia SOL 4.23

i.e.,

q2 = 0

and

q3 =- 0.1 rad

Correct option is (C). From the above solution, we have P2 = 0.1 P3 =- 0.2 since, P1 + P2 + P3 = 0 (Where P1 is injection at bus 1) So, P1 - P2 - P3 =- 0.1 + 0.2 = 0.1 pu Now, the apparent power delivered to base is 3 2 2 ^100 # 10 h V = S = 100 R 6 = 100 # 10 VA Therefore, the real power delivered by slack bus (bus 1) P = P1 S = ^0.1h # 100 # 106


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Power Systems

Chapter 4

= 10 # 106 watt = 10 MW SOL 4.24

Correct option is (B). For bus admittance matrix, Y11 + (Y12 + yline) + Y13 = 0 - j13 + (j10 + yline) + j 5 = 0 yline =- j2 Magnitude of susceptance is + 2

SOL 4.25

Correct option is (A). i1 (t) = Im sin (wt - f1) i2 (t) = Im cos (wt - f2)

nodia We know that, So, i1 (t) can be written as

cos (q - 90c) = sin q

i1 (t) = Im cos (wt - f1 - 90c) i2 (t) = Im cos (wt - f2) I1 = Im f1 + 90c

Now, in phasor form

I 2 = Im f 2 Current are balanced if I1 + I 2 = 0 Im f1 + 90c + Im f 2 = 0 Im cos ^f1 + 90ch + jIm sin ^f1 + 90ch + cos f 2 + j sin f 2 = 0 Im 8cos ^f1 + 90ch + j sin ^f1 + 90chB + Im 6cos f 2 + j sin f 2@ = 0 Im 8cos ^f1 + 90ch + cos f 2B + jIm 8sin f 2 + sin ^f1 + 90chB = 0 cos ^f1 + 90ch + cos f2 = 0 cos ^f1 + 90ch =- cos f2 = cos ^p + f2h f1 + 90c = p + f2 f1 = p + f2 2

or, SOL 4.26

Correct option is (A). Let penalty factor of plant G , is L1 given as 1 L1 = 1 - 2PL 2PG

1

2 G1

PL = 0.5P 2PL = 0.5 (2P ) = P G G 2PG 1 L1 = 1 - PG 1

1

2

So,

2

Penalty factor of plant G2 is L2 = For economic power generation

1 =1 2 1 - PL 2PG

2PL ca 2PG = 0 m 2

2

C1 # L1 = C 2 # L 2 where C1 and C2 are the incremental fuel cost of plant G1 and G2 .


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Power Systems

(10000) b

So,

Page 245

1 = 12500 # 1 1 - PG l 4 = 1-P G 5 PG = 1 pu 5 PG = 1 # 100 = 20 MW 5 2 PL = 0.5 b 1 l = 1 pu 5 50 PL = 1 # 100 = 2 MW 50 2

2

2

It is an 100 MVA, so

2

Loss or

PL = PG + PG - PL 40 = 20 + P2 - 2

Total power,

1

2

nodia PG = 22 MW 2

SOL 4.27

Correct option is (C). For double line-to-ground (LLG ) fault, relation between sequence current is I positive =-Î negative + I zeroh Gives values satisfy this relation, therefore the type of fault is LLG .

SOL 4.28

Correct option is (B). Complex power for generator

SG = SD1 + SD2 = 1 + 1 = 2 pu (Line is lossless) Power transferred from bus 1 to bus 2 is 1 pu, so V1 V2 sin (q1 - q2) 1= X V1 = V2 = 1 pu = 1 # 1 sin (q1 - q2) 0.5 X = 0.5 pu 0.5 = sin (q1 - q2) q1 - q2 = 30c q2 = q1 - 30c =- 30c (q1 = 0c) So, V1 = 1 0c V 1

V2 = 1 - 30c V 1 0c - 1 30c I12 = V1 - V2 = j 0.5 Z

Current,

= (1 - j 0.288) pu Current in SD is I2 , 2

SD = V2 I2) 1 = 1 - 30c I2) 2

I2 = 1 - 30c pu IG = I2 - I12 = 1 - 30c - (1 - j 0.288)

Current in QG , 2

= 0.268 - 120c VAR rating of capacitor, QC = V2 VG = 1 # 0.268 = 0.268 pu SOL 4.29



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Power Systems

Chapter 4

Total reactance, X = j1 + j 0.5 = j1.5 pu Critical angle is given as, dcr = cos-1 [(p - 2d0) sin d0 - cos d0] d0 " steady state torque angle. Steady state power is given as Pm = Pmax sin d0 E V Pmax = X E V Pm = sin d0 X (1.5) (1) sin d0 0.5 = 1.5

where,

nodia So,

...(i)

Pm = 0.5 pu

sin d0 = 0.5 d0 = 30c

d0 = 30c # p = 0.523 180c

In radian,

Substituting d0 into equation (i)

dcr = cos-1 [(p - 2 # 0.523) sin 30c - cos 30c] = cos-1 [(2.095) (0.5) - 0.866] = cos-1 (0.1815) - 79.6c

SOL 4.30


SOL 4.31

Correct option is (A). Negative phase sequence relay is used for the protection of alternators against unbalanced loading that may arise due to phase-to-phase faults.

SOL 4.32

Correct option is (C). Steady state stability or power transfer capability E V Pmax = X To improve steady state limit, reactance X should be reduced. The stability may be increased by using two parallel lines. Series capacitor can also be used to get a better regulation and to increase the stability limit by decreasing reactance. Hence (C) is correct option.

SOL 4.33

Correct option is (A). We know that loss \ PG2 loss \ length Distance of load from G1 is 25 km Distance of load from G2 & G 3 is 75 km generally we supply load from nearest generator. So maximum of load should be


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Power Systems

Page 247

supplied from G1 . But G2 & G 3 should be operated at same minimum generation. SOL 4.34

Correct option is (B). Power angle for salient pole alternator is given by V sin f + Ia Xq tan d = t Vt cos f + Ia Ra Since the alternator is delivering at rated kVA rated voltage Ia = 1 pu Vt = 1 pu f = 0c sin f = 0, cos f = 1 Xq = 1 pu, Xd = 1.2 pu 1 0 + 1 (1) tan d = # 1+0

nodia = 1 & d = 45c

SOL 4.35

Correct option is (B). The admittance diagram is shown below

Here

y10 =- 10j, y12 =- 5j, y23 = 12.5j, y 30 =- 10j

Note: y23 is taken positive because it is capacitive. Y11 = y10 + y12 =- 10j - 5j =- 15j Y12 = Y21 =- y21 = 5j Y13 = Y31 =- y13 = 0 Y22 = y20 + y21 + y23 = 0 + (- 5j) + (12.5j) = 7.5j Y23 = Y32 =- y23 =- 12.5j Y33 = y 30 + y13 + y23 =- 10j + 0 + 12.5j = 2.5j So the admittance matrix is RY Y Y V R- 15j 0 VW 5j S 11 12 13W S Y = SY21 Y22 Y33W = S 5j 7.5j - 12.5j W SSY Y Y WW SS 0 - 12.5j 2.5j WW 31 32 33 T X T X SOL 4.36

Correct option is (A). For generator G1 X mG = 0.25 # 100 = 0.1 pu 250 1

For generator G2

X mG = 0.10 # 100 = 0.1 pu 100 1

XL = XL = 0.225 # 10 = 2.25 W 2

1


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Power Systems

Chapter 4

For transmission lines L1 and L2 2 Z base = kV base = 15 # 15 = 2.25 W 100 MVA base X mL (pu) = 2.25 = 1 pu 2.25 X mL (pu) = 2.25 = 1 pu 2.25 So the equivalent pu reactance diagram 2

1

SOL 4.37

nodia

Correct option is (D). We can see that at the bus 3, equivalent thevenin’s impedance is given by Xth = ^ j0.1 + j1.0h || ^ j0.1 + j1.0h

= j1.1 || j1.1 = j0.55 pu Fault MVA = Base MVA Xth = 100 = 181.82 MVA 0.55

SOL 4.38

Correct option is (C). Given that, a a I > 0 so

SOL 4.39

I >0 VAB > 0 since it is Rectifier O/P VCD > 0 since it is Inverter I/P VAB > VCD , Than current will flow in given direction.

Correct option is (A). Given step voltage travel along lossless transmission line.

Voltage wave terminated at reactor as.


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Power Systems

Page 249

Applying KVL V + VL = 0 VL =- V VL =- 1 pu SOL 4.40

Correct option is (A). Given two buses connected by an Impedance of (0 + j5) W The Bus ‘1’ voltage is 100+30c V and Bus ‘2’ voltage is 100+0c V We have to calculate real and reactive power supply by bus ‘1’

nodia P + jQ = VI)

= 100+30c ;100+30c - 100+0cE 5j

= 100+30c [20+ - 60c - 20+ - 90c] = 2000+ - 30c - 2000+ - 60c P + jQ = 1035+15c P = 1035 cos 15c = 1000 W Q = 1035 sin 15c = 268 VAR

Real power

Reactive power

SOL 4.41

Correct option is (C). Given 3-f, 33 kV oil circuit breaker. Rating 1200 A, 2000 MVA, 3 sec Symmetrical breaking current Ib = ? Ib = MVA kA 3 kV = 2000 = 34.99 kA - 35 kA 3 # 33

SOL 4.42

Correct option is (C). Given a stator winding of an alternator with high internal resistance fault as shown in figure

Current through operating coil I1 = 220 # 5 A, I2 = 250 # 5 A 400 400 Operating coil current

= I 2 - I1


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Power Systems

Chapter 4

= (250 - 220) # 5/400 = 0.375 Amp SOL 4.43

Correct option is (C). Zero sequence circuit of 3-f transformer shown in figure is as following:

nodia No option seems to be appropriate but (C) is the nearest.

SOL 4.44

SOL 4.45

Correct option is (D). Given that A 50 Hz Generator is initially connected to a long lossless transmission line which is open circuited as receiving end as shown in figure. Due to ferranti effect the magnitude of terminal voltage does not change, and the field current decreases. Correct option is (B). Given : 3-f, 50 Hz, 11 kV distribution system, We have to find out e1, e2 = ? Equivalent circuit is as following

11 (6C) e1 = 3 = 11 # 6 = 3.46 kV 11 6C + 5C 3 e2 = 11 # 5 11 3 = 2.89 kV SOL 4.46

Correct option is (A). Given : 3-f, 50 Hz, 11 kV cable C1 = 0.2 mF C2 = 0.4 mF Charging current IC per phase = ? Capacitance Per Phase, C = 3C1 + C2 C = 3 # 0.2 + 0.4 = 1 mF


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Power Systems

Page 251

w = 2pf = 314 Changing current IC = V = V (wC) XC 3

= 11 # 10 # 314 # 1 # 10- 6 3 = 2 Amp SOL 4.47

Correct option is (B). Generator G1 and G2 XG1 = XG2 = X old # New MVA # b Old kV l New kV Old MVA 2 = j0.9 # 200 # b 25 l = j0.18 100 25 2 Same as XT1 = j0.12 # 200 # b 25 l = j0.27 90 25 2 XT2 = j0.12 # 200 # b 25 l = j0.27 90 25 X Line = 150 # 220 2 = j0.62 (220)

2

nodia The Impedance diagram is being given by as

SOL 4.48


SOL 4.49

Correct option is (C). We know complex power

a

So SOL 4.50

S = P + jQ = VI (cos f + j sin f) = VIe jf I = S jf Ve Real Power loss = I2 R 2 2 2 1 a S j2R = Constant PL = c S jf m R = S j2R f # 2 V Ve e e f PL \ 12 V

Correct option is (C). YBus matrix of Y-Bus system are given as V R S- 5 2 2.5 0 W S 2 - 10 2.5 0 W YBus = j S W S2.5 2.5 - 9 4 W 4 4 - 8W S0 X T We have to find out the buses having shunt element


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Power Systems

We know Here

Chapter 4

V R Sy11 y12 y13 y14W Sy21 y22 y23 y24W YBus = S W Sy 31 y 32 y 33 y 34W Sy 41 y 42 y 43 y 44W X T y11 = y10 + y12 + y13 + y14 =- 5j y22 = y20 + y21 + y23 + y24 =- 10j y 33 = y 30 + y 31 + y 32 + y 34 =- 9j y 44 = y 40 + y 41 + y 42 + y 43 =- 8j y12 = y21 =- y12 = 2j y13 = y 31 =- y13 = 2.5j y14 = y 41 =- y14 = 0j y23 = y 32 =- y23 = 2.5j y24 = y 42 =- y24 = 4j y10 = y11 - y12 - y13 - y14 =- 5j + 2j + 2.5j + 0j =- 0.5j

nodia So

y20 = y22 - y12 - y23 - y24 =- 10j + 2j + 2.5j + 4j =- 1.5j y 30 = y 33 - y 31 - y 32 - y 34 =- 9j + 2.5j + 2.5j + 4j = 0 y 40 = y 44 - y 41 - y 42 - y 43 =- 8j - 0 + 4j + 4j = 0 Admittance diagram is being made by as

From figure. it is cleared that branch (1) & (2) behaves like shunt element. SOL 4.51

Correct option is (B). We know that • Shunt Capacitors are used for improving power factor. • • •

SOL 4.52

Series Reactors are used to reduce the current ripples. For increasing the power flow in line we use series capacitor. Shunt reactors are used to reduce the Ferranti effect.

Correct option is (C). We know that for different type of transmission line different type of distance relays are used which are as follows. Short Transmission line -Ohm reactance used Medium Transmission Line -Reactance relay is used


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Power Systems

Page 253

Long Transmission line -Mho relay is used SOL 4.53

Correct option is (C). Given that three generators are feeding a load of 100 MW. For increased load power demand, Generator having better regulation share More power, so Generator -1 will share More power than Generator -2.

SOL 4.54

Correct option is (A). Given Synchronous generator of 500 MW, 21 kV, 50 Hz, 3-f, 2-pole P.F = 0.9 , Moment of inertia M = 27.5 # 103 kg-m2 Inertia constant H = ? Generator rating in MVA G = P = 500 MW = 555.56 MVA 0.9 cos f N = 120 # f = 120 # 50 = 3000 rpm 2 pole 2 Stored K.E = 1 Mw2 = 1 M b 2pN l 2 2 60 = 1 # 27.5 # 103 # b 2p # 3000 l MJ 2 60

nodia = 1357.07 MJ Stored K.E Inertia constant (H) = Rating of Generator (MVA) H = 1357.07 555.56 = 2.44 sec SOL 4.55

Correct option is (D). Given for X to F section of phase ‘a’ Va -Phase voltage and Ia -phase current. Impedance measured by ground distance, Bus voltage = Va W Relay at X = Ia Current from phase 'a'

SOL 4.56

Correct option is (D). For EHV line given data is Length of line = 300 km and b = 0.00127 S rad/km wavelength l = 2p = 2p = 4947.39 km 0.00127 b l % = 300 So 100 = 0.06063 # 100 4947.39 # l l % = 6.063 l

SOL 4.57

Correct option is (B). For three phase transmission line by solving the given equation VRI V RDV V R(X - X ) 0 0 m WS aW S aW S s We get, (Xs - Xm) 0 0 WSIbW SDVbW = S SSDV WW SS (Xs + 2Xm)WWSSIcWW 0 0 c XT X T X T Zero sequence Impedance = Xs + 2Xm = 48

...(1)


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Power Systems

Chapter 4

and Positive Sequence Impedance = Negative Sequence Impedance = (Xs - Xm) = 15

...(2)

By solving equation (1) and (2) Zs or Xs = 26 and Zm or Xm = 11 SOL 4.58


SOL 4.59

Correct option is (B). SIL has no effect of compensation So

SOL 4.60

SIL = 2280 MW

nodia Correct option is (C). Given and

from equation (2)

PG1 + PG2 = 250 MW C1 (PG1) = PG1 + 0.055PG12 4 C2 (PG2) = 3PG2 + 0.03PG22

dC1 = 1 + 0.11P G1 dPG1 dC2 = 3 + 0.06P and G2 dPG2 Since the system is loss-less dC1 = dC2 Therefore dPG2 dPG1 So from equations (3a) and (3b) We have 0.11PG1 - 0.06PG2 = 2 Now solving equation (1) and (4), we get

...(1) ...(2)

...(3a) ...(3b)

...(4)

PG1 = 100 MW PG2 = 150 MW SOL 4.61

Correct option is (B). After connecting both the generators in parallel and scheduled to supply 0.5 Pu of power results the increase in the current. ` Critical clearing time will reduced from 0.28 s but will not be less than 0.14 s for transient stability purpose.

SOL 4.62

Correct option is (D). Given that the each section has equal impedance. Let it be R or Z , then by using the formula line losses = / I2 R On removing (e1); losses = (1) 2 R + (1 + 2) 2 R + (1 + 2 + 5) 2 R = R + 9R + 64R = 74R Similarly, On removing e2 ;losses = 52 R + (5 + 2) 2 R + (5 + 2 + 1) 2 R = 138R lossess on removing e 3 = (1) 2 R + (2) 2 R + (5 + 2) 2 R = 1R + 4R + 49R = 54R


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Power Systems

Page 255

on removing e 4 lossless = (2) 2 R + (2 + 1) 2 R + 52 R = 4R + 9R + 25R = 38R So, minimum losses are gained by removing e 4 branch. SOL 4.63

Correct option is (A). Given : V (t) = Vm cos (wt) For symmetrical 3 - f fault, current after the fault 2 Vm cos (wt - a) Z At the instant of fault i.e t = t 0 , the total current i (t) = 0 i (t) = Ae- (R/L) t +

0 = Ae- (R/L) t + 0

`

2 Vm cos (wt - a) 0 Z

nodia Ae- (R/L) t =- 2 Vm cos (wt 0 - a) Z Maximum value of the dc offset current 0

Ae- (R/L) t =- 2 Vm cos (wt 0 - a) Z 0

For this to be negative max.

(wt 0 - a) = 0 t0 = a w

or

...(1)

Z = 0.004 + j0.04

and

Z = Z +a = 0.0401995+84.29c a = 84.29cor 1.471 rad. From equation (1) t0 =

1.471 = 0.00468 sec (2p # 50) t 0 = 4.682 ms

SOL 4.64

Correct option is (C). Since the fault ‘F’ is at mid point of the system, therefore impedance seen is same from both sides.

Z = 0.0201+84.29c 2 Z1 (Positive sequence) = Z = 0.0201+84.29c 2 also Z1 = Z2 = Z 0 (for 3-f fault) `

1+0c I f (pu) = 1+0c = Z1 0.0201+84.29c

So magnitude

If

(p.u.)

= 49.8


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Power Systems

I f = 49.8 #

` Fault current SOL 4.65

Chapter 4

100 = 7.18 kA 3 # 400

Correct option is (A). If fault is LG in phase ‘a’

nodia Z1 = Z = 0.0201+84.29c 2

and

Then

and

So

Z2 = Z1 = 0.0201+84.29c Z 0 = 3Z1 = 0.0603+84.29c Ia /3 = Ia1 = Ia2 = Ia0 Ia1 (pu) = 1.0+0c Z1 + Z 2 + Z 0 1 .0 Ia1 = = 9.95 pu (0.0201 + 0.0201 + 0.0603)

Fault Current I f = Ia = 3Ia1 = 29.85 pu 100 Fault current I f = 29.85 # = 4.97 kA 3 # 400

SOL 4.66

Correct option is (A). a Equal Phase shift of point A & B with respect to source from both bus paths. So the type of transformer Y-Y with angle 0c.

SOL 4.67

Correct option is (C). Given incremental cost curve

PA + PB = 700 MW


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Power Systems

Page 257

For optimum generator PA = ? , PB = ? From curve, maximum incremental cost for generator A = 600 at 450 MW and maximum incremental cost for generator B = 800 at 400 MW minimum incremental cost for generator B = 650 at 150 MW Maximim incremental cost of generation A is less than the minimum incremental constant of generator B. So generator A operate at its maximum load = 450 MW for optimum generation. PA = 450 MW PB = (700 - 450) = 250 MW

SOL 4.68

SOL 4.69

nodia

Correct option is (C). Here power sharing between the AC line and HVDC link can be changed by controlling the HVDC converter alone because before changing only grid angle we can change the power sharing between the AC line and HVDC link. Correct option is (B). We have to find out maximum electric field intensity at various points. Electric field intensity is being given by as follows

From figures it is cleared that at point Y there is minimum chances of cancelation. So maximum electric field intensity is at point Y. SOL 4.70

Correct option is (D). To increase capacitive dc voltage slowly to a new steady state value first we have to make d =- ve than we have to reach its original value.


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Power Systems

Chapter 4

Correct option is (B). Given that .045 2p # 50 1 1 = 1.2 pu & C = 2p # 50 # 1.2 = 0.045 pu & L =

Reactance of line Suspectance of Line

Velocity of wave propagation = 3 # 105 Km/sec Length of line l = ? We know velocity of wave propagation VX = l LC l = VX LC .45 1 1 = 185 Km = 3 # 105 2p # 50 # 2p # 50 # 1.2

SOL 4.72

SOL 4.73

nodia

Correct option is (C). Due to the fault ‘F’ at the mid point and the failure of circuit-breaker ‘4’ the sequence of circuit-breaker operation will be 5, 6, 7, 3, 1, 2 (as given in options) (due to the fault in the particular zone, relay of that particular zone must operate first to break the circuit, then the back-up protection applied if any failure occurs.) Correct option is (A).

Solving we get

V R 1 - 13 W RSiaVW S 0 3 1 W R = [Van Vbn Vcn] S- 13 0 SibW 3 WS W S 1 1 0 W SicW S 3 - 3 XT X T

R = ;Van (ib - ic) + Vbn (ic - ia) + Vc (ia - ib)E 3 3 3 (i - i ) R = 3 (VI) , where b c = I and Van = V 3 SOL 4.74


Here

Since

P1 " power before the tripping of one ckt P2 " Power after tripping of one ckt P = EV sin d X Pmax = EV X


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Power Systems

P2 max = EX , X2

`

Page 259

here, [X2 = (0.1 + X) (pu)]

To find maximum value of X for which system does not loose synchronism P2 = Pm (shown in above figure) EV sin d = P ` m 2 X2 as Pm = 1 pu, E = 1.0 pu,V = 1.0 pu 1.0 # 1.0 sin 130c = 1 X2 & & &

SOL 4.75

X2 = 0.77 (0.1 + X) = 0.77 X = 0.67

nodia Correct option is (B). Given that

FP = KAFS

Rf V Rf V S aW S pW where, Phase component FP = SfbW, sequence component FS = SfnW SSf WW SSf WW c o T X T X R 1 1 1V S W and A = Sa2 a 1W SS a a2 1WW VP = TKAVS X ` 3 IP = KAIS and VS = Zl [IS ] R0.5 0 0 V S W where Zl = S 0 0.5 0 W SS 0 0 2.0WW T X We have to find out Z if VP = ZIP From equation (2) and (3) VP = KAZl [IS ] -1 VP = KAZlb A l I p K VP = AZlA- 1 I p R 1 1 1V S W A = Sa2 a 1W SS a a2 1WW T X Adj A -1 A = A R 2V S1 a a W Adj A = S1 a2 a W S W S1 1 1 W T X 1 A = 3 R 2V S1 a a W A- 1 = 1 S1 a2 a W 3S W S1 1 1 W T X

...(1)

...(2) ...(3)

...(4)

...(5)


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Power Systems

Chapter 4

From equation (5)

SOL 4.76

R 1 0.5 0.5V R 1 1 1VR0.5 0 0VR1 a a2V W S W W S W S S Vp = 1 Sa2 a 1WS 0 0.5 0WS1 a2 a W I p = S0.5 1 0.5W I p ...(6) 3S W S SS0.5 0.5 1 WW S a a2 1WWSS 0 0 2WWS1 1 1 W XT XT T X T and (6) X Comparing of equation (5) R 1 0.5 0.5V S W Z = S0.5 1 0.5W SS0.5 0.5 1 WW T X Correct option is (A). Given that the first two power system are not connected and separately loaded. Now these are connected by short transmission line. as P1 = P2 = Q1 = Q2 = 0 So here no energy transfer. The bus bar voltage and phase angle of each system should be same than angle difference is

nodia q = 30c - 20c = 10c

SOL 4.77

Correct option is (B). Given that, 230 V, 50 Hz, 3-f, 4-wire system P = Load = 4 kw at unity Power factor IN = 0 through the use of pure inductor and capacitor Than L = ?, C = ?

a IN = 0 = IA + IB + IC Network and its Phasor is being as

...(1)

Here the inductor is in phase B and capacitor is in Phase C. We know P = VI 3 So Ia = P = 4 # 10 = 17.39 Amp. 230 V From equation (1) IA =- (IB + IC ) `

IA =-c IB #

`

IA =

Now and

a Ib - Ic

3 +I 3 C # 2 2 m

3 IB = 3 IC IB - IC = 17.39 - 10 Amp 3 XC = V = 230 - 23 W 10 IC XC = 1 2pfC


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Power Systems

&

& So

SOL 4.78

Page 261

1 = 1 = 139.02 mF 2p # 50 # 23 2pfXC XL = V = 230 - 23 W = 2pfL 10 IL 23 = 72.95 mH L = XL = 2p # 100 2pf C =

L = 72.95 mH in phase B C = 139.02 mF in phase C

Correct option is (A). Maximum continuous power limit of its prime mover with speed governor of 5% droop. Generator feeded to three loads of 4 MW each at 50 Hz. Now one load Permanently tripped ` f = 48 Hz If additional load of 3.5 MW is connected than f = ? a Change in Frequency w.r.t to power is given as drop out frequency Df = # Change in power rated power = 5 # 3.5 = 1.16% = 1.16 # 50 = 0.58 Hz 15 100

nodia System frequency is = 50 - 0.58 = 49.42 Hz

SOL 4.79

Correct option is (B). With the help of physical length of line, we can recognize line as short, medium and long line.

SOL 4.80

Correct option is (A). For capacitor bank switching vacuum circuit breaker is best suited in view of cost and effectiveness.

SOL 4.81

Correct option is (B). Ratio of operating coil current to restraining coil current is known as bias in biased differential relay.

SOL 4.82

Correct option is (B). HVDC links consist of rectifier, inverter, transmission lines etc, where rectifier consumes reactive power from connected AC system and the inverter supplies power to connected AC system.

SOL 4.83

Correct option is (C). Given ABCD constant of 220 kV line A = D = 0.94+10c, B = 130+730c, C = 0.001+900c, VS = 240 kV % voltage regulation is being given as (V ) - (VR) Full load %V.R. = R No Load # 100 VR (Full load) At no load IR = 0 (VR) NL = VS /A , (VR) Full load = 220 kV


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Power Systems

Chapter 4

240 - 220 0 %V.R. = .94 # 100 220 %V.R. = 16 SOL 4.84


SOL 4.85

Correct option is (B). Given that, Vab1 = X+q1 , Vab2 = Y+q2 , Phase to neutral sequence volt = ? First we draw phasor of positive sequence and negative sequence.

nodia

From figure we conclude that positive sequence line voltage leads phase voltage by 30c VAN1 = X+q1 - 30c VAN2 = 4+q2 + 30c

SOL 4.86

Correct option is (A). For system base value 10 MVA, 69 kV, Load in pu(Z new ) = ? (MVA) old kVnew 2 Z new = Z old # # b kVold l (MVA) new 2 Z new = 0.72 # 20 # b 69 l = 36 pu 10 13.8

SOL 4.87

Correct option is (A). Unreliable convergence is the main disadvantage of gauss seidel load flow method.

SOL 4.88

Correct option is (C). Generator feeds power to infinite bus through double circuit line 3-f fault at middle of line. Infinite bus voltage(V ) = 1 pu Transient internal voltage of generator(E ) = 1.1 pu Equivalent transfer admittance during fault = 0.8 pu = 1/X delivering power(PS ) = 1.0 pu Perior to fault rotor Power angle d = 30c, f = 50 Hz Initial accelerating power(Pa ) = ? Pa = PS - Pm2 sin d = 1 - EV sin 30c X


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Power Systems

Page 263

= 1 - 1.1 # 1 # 1 = 0.56 pu 2 1/0.8 SOL 4.89

Correct option is (B). If initial acceleration power = X pu Initial acceleration = ? Inertia constant = ? X (pu) # S 180 # 50 # X # S = a = Pa = M S#S SH/180F a = 1800X deg / sec2 Inertia const. = 1 = 0.056 18

SOL 4.90


The post fault voltage at bus 1 and 3 are. Pre fault voltage. RV V R1+0cV S 1W S W VBus = SV2W = S1+0cW SSV WW SS1+0cWW 3 T X T X At bus 2 solid fault occurs Z (f) = 0 , r = 2 Fault current I f = Vr c = V2 c Zrr + Z f Z22 + Z f Z f = 1+0c =- 4j j0.24

Vi c = Prefault voltage Vi (f) = Vi c (0) - Zir I (f), V1 (f) = Vi c - Z12 I f = 1+0c - j0.08 (- j4) = 1 - 0.32 V1 (f) = 0.68 pu V3 (f) = V3 c - Z 32 I f = 1+0c - j0.16 (- j4) = 1 - 0.64 V3 (f) = 0.36 pu SOL 4.91


SOL 4.92

Correct option is (D). Rating of D-connected capacitor bank for unity p.f. PL = S cos f = 12 3 # 0.8 = 16.627 kW reactive power QL = S sin f = 12 3 # 0.6 = 12.47 kW For setting of unity p.f. we have to set capacitor bank equal to reactive power = 12.47 kW real power

SOL 4.93



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Power Systems

Chapter 4

Given that pu parameters of 500 MVA machine are as following M = 20 pu, X = 2 pu Now value of M and X at 100 MVA base are for inertia (M) (pu) new = (pu) old # old MVA new MVA (M pu) new = (M Pu) old # 500 100 = 20 # 5 = 100 pu 1 and for reactance (X ) (pu) new = (pu) old # new MVA old MVA (X pu) new = (X pu) old # 100 500 (X Pu) new = 2 # 1 = 0.4 pu 5

SOL 4.94

nodia Correct option is (D). 800 kV has Power transfer capacity = P At 400 kV Power transfer capacity = ? We know Power transfer capacity P = EV sin d X P \ V2 So if V is half than Power transfer capacity is 1 of previous value. 4

SOL 4.95

Correct option is (B). In EHV lines the insulation strength of line is governed by the switching over voltages.

SOL 4.96

Correct option is (A). For bulk power transmission over very long distance HVDC transmission preferably used.

SOL 4.97

Correct option is (D). Parameters of transposed overhead transmission line XS = 0.4 W/km , Xm = 0.1 W/km + ve sequence reactance X1 = ? Zero sequence reactance X0 = ? We know for transposed overhead transmission line. + ve sequence component X1 = X S - X m = 0.4 - 0.1 = 0.3 W/km Zero sequence component

SOL 4.98

X 0 = XS + 2Xm = 0.4 + 2 (0.1) = 0.6 W/km



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Power Systems

Page 265

Industrial substation of 4 MW load = PL QC = 2 MVAR for load p.f. = 0.97 lagging If capacitor goes out of service than load p.f. = ? cos f = 0.97 tan f = tan (cos- 1 0.97) = 0.25 QL - QC = 0.25 PL QL - 2 = 0.25 & QL = 3 MVAR 4 Q f = tan- 1 c L m = tan- 1 b 3 l = 36c 4 PL cos f = cos 36c = 0.8 lagging SOL 4.99


Y22 = ? I1 = V1 Y11 + (V1 - V2) Y12

= 0.05V1 - j10 (V1 - V2) =- j9.95V1 + j10V2 I2 = (V2 - V1) Y21 + (V2 - V3) Y23

= j10V1 - j9.9V2 - j0.1V3 Y22 = Y11 + Y23 + Y2 =- j9.95 - j9.9 - 0.1j =- j19.95

SOL 4.100

Correct option is (C). F1 = a + bP1 + cP 12 Rs/hour F2 = a + bP2 + 2cP 22 Rs/hour For most economical operation P1 + P2 = 300 MW then P1, P2 = ? We know for most economical operation 2F1 = 2F2 2P1 2P2 2cP1 + b = 4cP2 + b P1 = 2P2 P1 + P2 = 300 From eq (1) and (2)

...(1) ...(2)

P1 = 200 MW , P2 = 100 MW SOL 4.101

Correct option is (B). A B V2 V1 We know that ABCD parameters > H => C DH >I1H I1 , C = I1 B = V1 I2 V = 0 V2 I = 0 V1 Z 1 + Z2 In figure = 1 C = V1 Z2 Z Z1 + Z 2 # 2 2

2


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Power Systems

or SOL 4.102

Chapter 4

1 = 40+ - 45c Z2 = 1 = C 0.025+45c

Correct option is (D). Given

Steady state stability Power Limit = 6.25 pu If one of double circuit is tripped than

nodia Steady state stability power limit = ? Pm1 = EV = 1 # 1 = 6.25 X 0.12 + X 2 1 = 6.25 0.12 + 0.5X

& X = 0.008 pu If one of double circuit tripped than Pm2 = EV X 1 = 1#1 = 0.12 + 0.08 0.12 + X Pm2 = 1 = 5 pu 0.2

SOL 4.103

Correct option is (D). Given data Substation Level = 220 kV 3-f fault level = 4000 MVA LG fault level = 5000 MVA Positive sequence reactance: 4000 Fault current I f = 3 # 220 X1 = Vph /I f =

SOL 4.104

220 3 4000 3 # 220

= 220 # 220 = 12.1 W 4000

Correct option is (B). Zero sequence Reactance X 0 = ? If =

5000 3 # 220

Ia1 = Ia2 = Ia0 =

If 5000 = 3 3 3 # 220

220

Vph 3 = 5000 Ia1 220 # 3 220 220 = 29.04 W # X1 + X 2 + X 0 = 3 # 5000 X1 + X 2 + X 0 =


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Power Systems

Page 267

X1 = X2 = 12.1 W X 0 = 29.04 - 12.1 - 12.1 = 4.84 W SOL 4.105

Correct option is (B). Instantaneous power supplied by 3-f ac supply to a balanced R-L load. P = Va Ia + Va Ib + Vc Ic = (Vm sin wt) Im sin (wt - f) + Vm sin (wt - 120c) Im sin (wt - 120c - f) + Vm sin (wt - 240c) Im sin (wt - 240c - f) = VI [cos f - cos (2wt - f) + cos f - cos (2wt - 240 - f) + cos f - cos (2wt + 240 - f)] ...(1) P = 3VI cos f

nodia

equation (1) implies that total instantaneous power is being constant.

SOL 4.106

SOL 4.107

Correct option is (C). In 3-f Power system, the rated voltage is being given by RMS value of line to line voltage. Correct option is (B).

In this figure the sequence is being given as RBY SOL 4.108

Correct option is (C). In thermal power plants, the pressure in the working fluid cycle is developed by the help to feed water pump.

SOL 4.109

Correct option is (A). Kaplan turbines are used for harnessing low variable waterheads because of high percentage of reaction and runner adjustable vanes.

SOL 4.110

Correct option is (B). MHO relay is the type of distance relay which is used to transmission line protection. MHO Relay has the property of being inherently directional.

SOL 4.111

Correct option is (C). Surge impedance of line is being given by as Z =

L = C

11 # 10- 3 11.68 # 10- 9

= 306.88 W Ideal power transfer capability


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Power Systems

Chapter 4

2 (800) 2 P =V = = 2085 MW 306.88 Z0

SOL 4.112

Correct option is (D). Given that, Power cable voltage = 110 kV C = 125 nF/km Dielectric loss tangent = tan d = 2 # 10- 4 Dielectric power loss = ? dielectric power loss is given by P = 2V2 wC tan d = 2 (110 # 103) 2 # 2pf # 125 # 10- 9 # 2 # 10- 4

nodia

= 2 (121 # 108 # 2 # 3.14 # 50 # 250 # 10- 13) = 189 W/

km

SOL 4.113

Correct option is (A). Given data Lightening stroke discharge impulse current of I = 10 kA Transmission line voltage = 400 kV Impedance of line Z = 250 W Magnitude of transient over-voltage = ? The impulse current will be equally divided in both directions since there is equal distribution on both sides. Then magnitude of transient over-voltage is V = IZ/2 = 10 # 103 # 250 2 = 1250 # 103 V = 1250 kV

SOL 4.114

Correct option is (C). The A, B, C, D parameters of line A = D = 0.936+0.98c B = 142+76.4c C = (- 5.18 + j914) 10- 6 W At receiving end PR = 50 MW , VR = 220 kV p.f = 0.9 lagging VS = ? Power at receiving end is being given by as follows VS VR A VR 2 cos (b - d) cos (b - a) PR = B B VS # 220 0.936 (220) 2 = cos (76.4c - d) cos 75.6c 142 142 VS cos (76.4 - d) = 50 # 142 + 0.936 # 220 # 0.2486 220 = 32.27 + 51.19 VS cos (76.4 - d) = 83.46

...(1)


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Power Systems

Page 269

QR = PR tan f = PR tan (cos- 1 f) = 50 tan (cos- 1 0.9) = 24.21 MW VS VR A VR 2 sin (b - d) sin (b - a) QR = B B VS # 220 0.936 # (220) 2 = sin (76.4c - d) sin 75.6c 142 142 ...(2) (24.21) 142 + 0.936 # 220 # 0.9685 = VS sin (76.4c - d) 220 from equation (1) & (2)

Same as

VS

2

= (215) 2 + (83.46) 2

VS =

SOL 4.115

53190.5716 = 230.63 kV


A new generator of Eg = 1.4+30c pu XS = 1.0 pu, connected to bus of Vt Volt Existing Power system represented by thevenin’s equivalent as Eth = 0.9+0c, Zth = 0.25+90c, Vt = ?

From the circuit given Eg - Eth 1.212 + j7 - 0.9 = 1.4+30c - 0.9+0c = Zth + XS j (1.25) j (1.25) 0.312 + j7 = = 0.56 - 0.2496j j (1.25)

I =

Vt = Eg - IXS = 1.212 + j7 - (0.56 - 0.2496j) (j1) = 1.212 - 0.2496 + j (0.7 - 0.56) = 0.9624 + j0.14 Vt = 0.972+8.3c SOL 4.116

Correct option is (C). Given that 3-f Generator rated at 110 MVA, 11 kV Xdm = 19% , Xdl= 26% XS = 130% , Operating at no load 3-f short circuit fault between breaker and transformer symmetrical Irms at breaker = ? We know short circuit current Isc = 1 = 1 =- j5.26 pu Xdm j0.19 rating MVA of generator Base current IB = 3 # kV of generator


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Power Systems

Chapter 4

110 # 106 3 # 11 # 103 IB = 5773.67 Amp Symmetrical RMS current = IB # Isc = 5773.67 # 5.26 = 30369.50 Amp & Irms = 30.37 kA IB =

SOL 4.117

Correct option is (A). + ve sequence current Ia = 1 [Ia + aIb + a2 Ic] 3 = 1 [10+0c + 1+120c # 10+180c + 0] 3 = 1 [10+0c + 10+300c] = 1 [10 + 5 - j8.66] 3 3 = 1 [15 - j8.66] 3 = 17.32+ - 30c = 5.78+ - 30c 3

SOL 4.118

nodia

Correct option is (D). Given data 500 MVA , 50 Hz, 3 - f generator produces power at 22 kV Generator " Y connected with solid neutral Sequence reactance X1 = X2 = 0.15 , X 0 = 0.05 pu Sub transient line current = ? E 1 = = 1 =- 2.857j Ia1 = j0.15 + j0.15 + j0.05 0.35j Z1 + Z 2 + Z 0 Now sub transient Line current Ia = 3Ia1 Ia = 3 (- 2.857j) =- 8.57j

SOL 4.119

Correct option is (B). Given: 50 Hz, 4-Pole, 500 MVA, 22 kV generator p.f. = 0.8 lagging Fault occurs which reduces output by 40%. Accelerating torque = ? After fault,

Where

Power = 500 # 0.8 = 400 MW Power = 400 # 0.6 = 240 MW a Pa = Ta # w Ta = Pa w w = 2pfmechanical fmechanical = felectrical # 2 = felectrical # 2 4 P Pa = 400 - 240 = 160 MW 160 Ta = 2 # p # 50/2 Ta = 1.018 MN


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Power Systems

Page 271

Correct option is (D). N = 250 rpm f = 50 Hz.

Turbine rate speed To produce power at

P =? a N = 120 f P P = 120 f = 120 # 50 = 24 250 N

No. of Poles

P = 24 Poles SOL 4.121

SOL 4.122

Correct option is (C). In case of bundled conductors, We know that self GMD of conductor is increased and in a conductor critical disruptive voltage of line depends upon GMD of conductor. Since GMD of conductor is increased this causes critical disruptive voltage is being reduced and if critical disruptive voltage is reduced, the corona loss will also be reduced.


Given that no. of buses n = 300 Generator bus = 20 Reactive power support buses = 25

Fixed buses with Shunt Capacitor = 15 Slack buses (ns ) = 20 + 25 - 15 = 30 a Size of Jacobian Matrix is given as

= 2 (n - ns) # 2 (n - ns) = 2 (300 - 30) # 2 (300 - 30) = 540 # 540 SOL 4.123

Correct option is (B). Auxiliary component in HVDC transmission system are DC line inductor and reactive power sources.

SOL 4.124

Correct option is (C). a Exchanged electrical power is being given as follows P = EV 6sin (d1 - d2)@ Xd Given that

from eq(1),

P E V Xd

...(1)

" Power supply by generator = 0.5 pu " Voltage for rotar generator = 2.0 pu " Voltage of motor rotor = 1.3 pu = X eq = Reactance of generator + Reactance of motor + Recatance of connecting line

Xd = 1.1 + 1.2 + 0.5 = 2.8 d1 - d2 = Rotor angle difference = ? 0.5 = 2 # 1.3 sin (d1 - d2) 2.8


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Power Systems

Chapter 4

& d1 - d2 = sin- 1 b 2.8 # 0.5 l 2.6 = 32.58 SOL 4.125

Correct option is (B). Time period between energization of trip circuit and the arc extinction on an opening operation is known as the interrupting time of Circuit breaker.

SOL 4.126

Correct option is (B). Given that ABCD parameters of line as A = D = 0.9+0c, B = 200+90% W , C = 0.95 # 10 - 3 +90% S . at no-load condition,

nodia Receiving end voltage(VR) = sending end voltage(VS ) ohmic value of reactor = ? We know VS = AVR + BIR VS = VR VR = AVR + BIR VR (1 - A) = BIR VR = B = 200+90c IR 1-A 1 - 0.9+0c VR = 2000+90c IR The ohmic value of reactor = 2000 W

SOL 4.127

Correct option is (A). Surge impedance of cable Z1 =

L; C

L = 0.4 mH/km, C = 0.5 mF /km -3

0.4 # 10 = 28.284 = 0.5 # 10- 6 Surge impedance of overhead transmission line Z2 = Z3 = L ; L = 1.5 mm/km, C = 0.015 mF/km C 1.5 # 10- 5 = 316.23 0.015 # 10- 6 Now the magnitude of voltage at junction due to surge is being given by as V = 20 kV Vl = 2 # V # Z2 Z 2 + Z1 Z2 = Z 3 =

3 = 2 # 20 # 10 # 316.23 316 + 28.284

= 36.72 kV SOL 4.128

Correct option is (D). Let that current in line is I amp than from figure current in line section PR is (I - 10) amp current in line section RS is (I - 10 - 20) = (I - 30) amp current in SQ Section is (I - 30 - 30) = (I - 60) amp


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Power Systems

Page 273

Given that VP and VQ are such that VP - VQ = 3 V Applying KVL through whole line VP - VQ = (I - 10) 0.1 + (I - 30) 0.15 + (I - 60) # 0.2 & 3 = 0.45I - 17.5 I = 20.5 = 45.55 amp 0.45 Now the line drop is being given as = (I - 10) 0.1 + (I - 30) 0.15 + (I - 60) 0.2 = (33.55) 0.1 + (15.55) 0.15 + (14.45) 0.2 = 8.58 V The value of VP for minimum voltage of 220 V at any feeder is

nodia = 220 + Line voltage = 220 + 8.58 = 228.58 V

SOL 4.129

Correct option is (B). A balanced delta connected load = 8 + 6j = 2 V2 = 400 volt Improved Power Factor cos f2 = 0.9

f1 = tan- 1 ^6/8h = 36.85c f2 = cos- 1 (0.9) = 25.84c I = V = 400 8 + 6j Z 400 = = 40+ - 36.86c 10+36.86c = 32 - j24

Since Power factor is Improved by connecting a Y-connected capacitor bank like as

Phasor diagram is being given by as follows


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Power Systems

Chapter 4

oa = I l cos f2 = I cos f1

In figure

I l cos 25.84c = 32 I l # 0.9 = 32 Il = 35.55 ac = 24 Amp. ab = I l sin f2 = 35.55 sin 25.84c ab = 15.49 Amp

(ac = I sin f1)

Ic = bc = ac - ab = 24 - 15.49 = 8.51 Amp KVAR of Capacitor bank = 3 # V # IC = 3 # 400 # 8.51 1000 1000 = 10.2 KVAR SOL 4.130


Given Load Power = 100 MW VS = VR = 11 kV p.u. # (kV) 2 j0.2 # (11) 2 Impedance of line = ZL = = j0.242 W 100 MV VS VR sin d We know PL = X 3

3

100 # 106 = 11 # 10 # 11 # 10 sin d 0.242 100 # 0.242 = sin d 121

d = sin- 1 (0.2) = 11.537c Reactive Power is being given by VS VR VR 2 QL = cos d X X 3 2

3 3 (11 # 10 ) = 11 # 10 # 11 # 10 cos (11.537c) 0.242 0.242 6

= 121 # 10 [cos (11.537c) - 1] =- 10.1 MVAR 0.242 SOL 4.131

Correct option is (B). Given the bus Impedance Matrix of a 4-bus Power System V R Sj0.3435 j0.2860 j0.2723 j0.2277W Sj0.2860 j0.3408 j0.2586 j0.2414W Z bus = S W Sj0.2723 j0.2586 j0.2791 j0.2209W Sj0.2277 j0.2414 j0.2209 j0.2791W Now a branch osT j0.2 W is connected between bus X2 and reference RZ V ij 1 SSh WW Z g Z ZB(New) = ZB (Old) jnB Zij + Zb S W8 ji SZnjW T X in jth and reference bus j = 2 , n = 4 so New element Zb = j0.2 W is connected R V SZ12W SZ22W 1 S W Z Z Z Z Zij + Zb SZ23W 8 21 22 23 24B SZ24W T X


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Power Systems

V R Sj0.2860W Sj0.3408W 1 = W8j0.2860 j0.3408 j0.2586 j0.2414B S 6j (0.3408) + j0.2@ Sj0.2586W Sj0.2414W X T Given that we are required to change only Z22, Z23 j2 (0.3408) 2 So in equation (1) = j0.2147 Zl22 = j (0.5408) j2 (0.3408) (0.2586) = j0.16296 Zl23 = 0.5408

Page 275

...(1)

Z22(New) = Z22(Old) - Zl22 Z23(New)

= j0.3408 - j0.2147 = j0.1260 = Z23 (Old) - Zl23

nodia = j0.2586 - j0.16296 = j0.0956

SOL 4.132

Correct option is (D). Total zero sequence impedance, + ve sequence impedance and - ve sequence impedances Z 0 = (Z 0) Line + (Z 0) Generator = j0.04 + j0.3 = j0.34 pu Z1 = (Z1) Line + (Z1) Generator = j0.1 + j0.1 = j0.2 pu Z2 = (Z2) Line + (Z2) Generator = j0.1 + j0.1 = j0.2 pu Zn = j0.05 pu

for L-G fault

Ea Z 0 + Z 1 + Z 2 + 3Z n 0.1 = j0.2 + j0.2 + j0.34 + j0.15

Ia1 =

=- j1.12 pu generator MVA IB = 3 generator kV 20 # 106 = = 1750 Amp 3 # 6.6 # 103 Fault current I f = (3Ia) IB = 3 (- j1.12) (1750) =- j5897.6 Amp Neutral Voltage and

Vn = I f Zn Zn = ZB # Z pu (6.6) 2 = 0.05 = 0.1089 W 20 # Vn = 5897.6 # 0.1089 = 642.2 V

SOL 4.133

Correct option is (A). We know that Optimal Generation IC1 = IC2 , and P3 = 300 MW (maximum load) (Independent of load) IC 3 = 30 20 + 0.3P1 = 30 + 0.4 P2


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Power Systems

Chapter 4

0.3P1 - 0.4P2 = 10 P1 + P2 + P3 = 700 P1 + P2 + 300 = 700

...(1)

P1 + P2 = 400 From equation (1) and (2)

...(2)

P1 = 242.8 MW P2 = 157.14 MW SOL 4.134

SOL 4.135

Correct option is (A). For transmission line protection-distance relay For alternator protection-under frequency relay For bus bar protection-differential relay For transformer protection-Buchholz relay

nodia Correct option is (C).

We know by equal area criteria PS (dm - d0) =

#d

dm

Pmax sin ddd

C

Pmax sin d0 (dm - d0) = Pmax [cos d0 - cos dm] Pmax = 2 P0 = Pmax sin d0 = 1 d0 = 30c dmax = 110c (given) Now from equation (1) 2 sin 30c (110 - 30) p = 2 [cos dc - cos 110c] 180 0.5 # 80p = cos dc + 0.342 180

...(1)

cos dc = 0.698 - 0.342 dc = 69.138c SOL 4.136

Correct option is (D). a Both sides are granted So, Ia = Ea = 10+0c = 5+ - 90c 2j Za Ib = Eb = 10+ - 90c = 3.33+ - 180c 3j Zb Ic = Ec = 10+120c = 2.5+30c 4j Zc


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Power Systems

We know

Page 277

Ia = 1 [Ia + aIb + a2 Ic] 3 1

where a = 1+120c & a2 = 1+240c Ia1 = 1 [5+ - 90c + 3.33+ ^- 180c + 120ch + 2.5+ ^240c + 30ch] 3 Ia1 = 1 [5+ - 90c + 3.33+ - 60c + 2.5+270c] 3 = 1 [- 5j + 1.665 - j2.883 - 2.5j] 3 = 1 [1.665 - j10.383] = 3.5+ - 80.89c 3 ***********

nodia


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Control Systems

Chapter 5

CHAPTER 5 CONTROL SYSTEMS


ONE MARK

In the formation of Routh-Hurwitz array for a polynomial, all the elements of a row have zero values. This premature termination of the array indicates the presence of (A) only one root at the origin

nodia (B) imaginary roots

(C) only positive real roots (D) only negative real roots

EE SP 5.2

The root locus of a unity feedback system is shown in the figure.

The closed loop transfer function of the system is C ^s h C ^s h -K K (B) (A) = = R ^s h ^s + 1h^s + 2h + K R ^s h ^s + 1h^s + 2h C ^s h C ^s h K K (C) (D) = = R ^s h ^s + 1h^s + 2h - K R ^s h ^s + 1h^s + 2h + K YEAR 2014 EE01

TWO MARKS

EE SP 5.3

For the given system, it is desired that the system be stable. The minimum value of a for this condition is ______.

EE SP 5.4

The Bode magnitude plot of the transfer function K ^1 + 0.5s h^1 + as h G ^s h = s a1 + s k^1 + bs ha1 + s k 8 36 is shown below :


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Control Systems

Page 279

Note that - 6 dB/octave =- 20 dB/decade . .

nodia The value of a is _____ bK

YEAR 2014 EE02

EE SP 5.5

EE SP 5.6

ONE MARK

4 The closed-loop transfer function of a system is T ^s h = 2 . The + s 0 . 4s + 4 h ^ steady state error due to unit step input is _____.

The state transition matrix for the system 1 0 x1 1 xo1 > o H = >1 1H>x H + >1H u is x2 2 t e 0 et 0 (A) > t tH (B) > 2 t tH e e te e et 0 (C) > t tH te e

et tet H (D) > 0 et

YEAR 2014 EE02

TWO MARKS

EE SP 5.7

A system with the open loop transfer function K G ^s h = s ^s + 2h^s2 + 2s + 2h is connected in a negative feedback configuration with a feedback gain of unity. For the closed loop system to be marginally stable, the value of K is ______

EE SP 5.8

For the transfer function

5 ^s + 4h s ^s + 0.25h^s2 + 4s + 25h The values of the constant gain term and the highest corner frequency of the Bode plot respectively are (A) 3.2, 5.0 (B) 16.0, 4.0 (C) 3.2, 4.0 G ^s h =


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Control Systems

Chapter 5

(D) 16.0, 5.0 EE SP 5.9

The second order dynamic system dX = PX + Qu dt y = RX has the matrices P , Q and R as follows : -1 1 0 H Q = > H R = 60 1@ P => 0 -3 1 The system has the following controllability and observability properties: (A) Controllable and observable (B) Not controllable but observable (C) Controllable but not observable

nodia (D) Not controllable and not observable YEAR 2014 EE03

EE SP 5.10

ONE MARK

The signal flow graph of a system is shown below. U ^s h is the input and C ^s h is the output.

Assuming, h1 = b1 and h 0 = b 0 - b1 a1 , the input-output transfer function, C ^s h of the system is given by G ^s h = U ^s h (A) G ^s h = 2 b 0 s + b1 s + a 0 s + a1 (B) G ^s h = 2 a1 s + a 0 s + b1 s + b 0 (C) G ^s h = 2 b1 s + b 0 s + a1 s + a 0 (D) G ^s h = 2 a 0 s + a1 s + b 0 s + b1 EE SP 5.11

A single-input single output feedback system has forward transfer function G ^s h and feedback transfer function H ^s h. It is given that G ^s h H ^s h < 1. Which of the following is true about the stability of the system ? (A) The system is always stable (B) The system is stable if all zeros of G ^s h H ^s h are in left half of the s -plane (C) The system is stable if all poles of G ^s h H ^s h are in left half of the s -plane


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Control Systems

Page 281

(D) It is not possible to say whether or not the system is stable from the information given YEAR 2014 EE03 EE SP 5.12

TWO MARKS

The block diagram of a system is shown in the figure

If the desired transfer function of the system is C ^s h = 2 s R ^s h s +s+1 then G ^s h is (A) 1 (B) s

nodia (C) 1/s

EE SP 5.13

(D)

-s s3 + s2 - s - 2

Consider the system described by following state space equations 0 1 x1 0 x1 xo1 > o H = >- 1 - 1H>x H + >1H u ; y = 81 0B>x H x2 2 2

If u is unit step input, then the steady state error of the system is (A) 0 (B) 1/2 (C) 2/3 (D) 1

EE SP 5.14

The magnitude Bode plot of a network is shown in the figure

The maximum phase angle f m and the corresponding gain Gm respectively, are (A) - 30c and 1.73 dB (B) - 30c and 4.77 dB (C) + 30c and 4.77 dB (D) + 30c and 1.73 dB YEAR 2013 EE SP 5.15

ONE MARK

Assuming zero initial condition, the response y ^ t h of the system given below to a unit step input u ^ t h is


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Control Systems

(A) u ^ t h 2 (C) t u ^ t h 2 EE SP 5.16

Chapter 5

(B) tu ^ t h (D) e-t u ^ t h

The Bode plot of a transfer function G ^s h is shown in the figure below.

nodia

The gain _20 log G ^s h i is 32 dB and - 8 dB at 1 rad/s and 10 rad/s respectively. The phase is negative for all w. Then G ^s h is (A) 39.8 (B) 392.8 s s (C) 32 (D) 322 s s YEAR 2013

EE SP 5.17

Y ^s h The signal flow graph for a system is given below. The transfer function U ^s h for this system is

s+1 5s2 + 6s + 2 (C) 2 s + 1 s + 4s + 2 (A)

EE SP 5.18

TWO MARKS

s+1 s 2 + 6s + 2 (D) 2 1 5s + 6s + 2

(B)

The open-loop transfer function of a dc motor is given as w ^s h = 10 Va ^s h 1 + 10s When connected in feedback as shown below, the approximate value of Ka that will reduce the time constant of the closed loop system by one hundred times as


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Control Systems

Page 283

compared to that of the open-loop system is

(A) 1

(B) 5

(C) 10

(D) 100

Common Data For Q. 19 and 20 The state variable formulation of a system is given as 1 - 2 0 x1 x1 xo1 > H=> H> H + > H u , x1 ^0 h = 0 , x2 ^0 h = 0 and y = 61 0@> H 0 - 1 x2 1 x2 xo2

EE SP 5.19

nodia The response y ^ t h to the unit step input is (A) 1 - 1 e-2t (B) 1 - 1 e-2t - 1 e-t 2 2 2 2 (C) e-2t - e-t

EE SP 5.20

(D) 1 - e-t

The system is (A) controllable but not observable (B) not controllable but observable (C) both controllable and observable

(D) both not controllable and not observable YEAR 2012

TWO MARKS

EE SP 5.21

The state variable description of an LTI system is given by Jxo1N J 0 a1 0NJx1N J0N K O K OK O K O Kxo2O = K 0 0 a2OKx2O + K0O u Kxo O Ka 0 0OKx 3O K 1O 3 3 L P L PL P L P Jx1N K O y = _1 0 0iKx2O Kx 3O L P where y is the output and u is the input. The system is controllable for (B) a1 = 0, a2 ! 0, a 3 ! 0 (A) a1 ! 0, a2 = 0, a 3 ! 0 (C) a1 = 0, a 3 ! 0, a 3 = 0 (D) a1 ! 0, a2 ! 0, a 3 = 0

EE SP 5.22

The feedback system shown below oscillates at 2 rad/s when


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Control Systems

(A) K = 2 and a = 0.75 (C) K = 4 and a = 0.5

Chapter 5

(B) K = 3 and a = 0.75 (D) K = 2 and a = 0.5

Statement for Linked Answer Questions 23 and 24 : The transfer function of a compensator is given as Gc (s) = s + a s+b EE SP 5.23

Gc (s) is a lead compensator if (A) a = 1, b = 2 (B) a = 3, b = 2 (C) a =- 3, b =- 1 (D) a = 3, b = 1

EE SP 5.24

nodia The phase of the above lead compensator is maximum at (A) 2 rad/s (B)

3 rad/s

(C)

6 rad/s

(D) 1/ 3 rad/s

YEAR 2011

EE SP 5.25

The frequency response of a linear system G (jw) is provided in the tubular form below G (jw)

1.3

+G (jw) - 130c (A) 6 dB and 30c (C) - 6 dB and 30c EE SP 5.26

ONE MARK

1.2

1.0

0.8

0.5

0.3

- 140c

- 150c

- 160c

- 180c

- 200c

(B) 6 dB and - 30c (D) - 6 dB and - 30c

The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r (t) having a magnitude of 10 and a duration of one second, as shown in the figure is

(A) 0 (C) 1

(B) 0.1 (D) 10


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Control Systems

Page 285

An open loop system represented by the transfer function (s - 1) is G (s) = (s + 2) (s + 3) (A) Stable and of the minimum phase type (B) Stable and of the non–minimum phase type (C) Unstable and of the minimum phase type (D) Unstable and of non–minimum phase type YEAR 2011

EE SP 5.28

TWO MARKS

The open loop transfer function G (s) of a unity feedback control system is given as K bs + 2 l 3 G (s) = 2 s (s + 2) From the root locus, at can be inferred that when K tends to positive infinity, (A) Three roots with nearly equal real parts exist on the left half of the s -plane (B) One real root is found on the right half of the s -plane

nodia (C) The root loci cross the jw axis for a finite value of K; K ! 0 (D) Three real roots are found on the right half of the s -plane

EE SP 5.29

A two loop position control system is shown below

The gain K of the Tacho-generator influences mainly the (A) Peak overshoot (B) Natural frequency of oscillation (C) Phase shift of the closed loop transfer function at very low frequencies (w " 0) (D) Phase shift of the closed loop transfer function at very high frequencies (w " 3)

YEAR 2010

EE SP 5.30

TWO MARKS

o = AX + Bu with A = >- 1 2H, B = >0H is The system X 0 2 1 (A) Stable and controllable (B) Stable but uncontrollable (C) Unstable but controllable (D) Unstable and uncontrollable


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Control Systems

Chapter 5

The characteristic equation of a closed-loop system is s (s + 1) (s + 3) k (s + 2) = 0, k > 0 .Which of the following statements is true ? (A) Its root are always real (B) It cannot have a breakaway point in the range - 1 < Re [s] < 0 (C) Two of its roots tend to infinity along the asymptotes Re [s] =- 1 (D) It may have complex roots in the right half plane.

EE SP 5.32

1 plotted in the s (s + 1) (s + 2) complex G (jw) plane (for 0 < w < 3) is

The frequency response of G (s) =


EE SP 5.34

ONE MARK

The measurement system shown in the figure uses three sub-systems in cascade whose gains are specified as G1, G2, 1/G3 . The relative small errors associated with each respective subsystem G1, G2 and G3 are e1, e2 and e3 . The error associated with the output is :

(A) e1 + e2 + 1 e3

(B) e1 e2 e3

(C) e1 + e2 - e3

(D) e1 + e2 + e3

The first two rows of Routh’s tabulation of a third order equation are as follows. s3 2 2 s2 4 4 This means there are (A) Two roots at s = ! j and one root in right half s -plane (B) Two roots at s = ! j2 and one root in left half s -plane


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Control Systems

Page 287

(C) Two roots at s = ! j2 and one root in right half s -plane (D) Two roots at s = ! j and one root in left half s -plane EE SP 5.35

The polar plot of an open loop stable system is shown below. The closed loop system is

nodia (A) always stable

(B) marginally stable (C) un-stable with one pole on the RH s -plane (D) un-stable with two poles on the RH s -plane

EE SP 5.36

The asymptotic approximation of the log-magnitude v/s frequency plot of a system containing only real poles and zeros is shown. Its transfer function is

10 (s + 5) s (s + 2) (s + 25) 100 (s + 5) (C) s (s + 2) (s + 25) (A)

1000 (s + 5) s (s + 2) (s + 25) 80 (s + 5) (D) 2 s (s + 2) (s + 25) (B)

2


TWO MARKS

The unit-step response of a unity feed back system with open loop transfer function G (s) = K/ ((s + 1) (s + 2)) is shown in the figure. The value of K is

(A) 0.5 (C) 4

(B) 2 (D) 6


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Control Systems

Chapter 5

The open loop transfer function of a unity feed back system is given by G (s) = (e - 0.1s) /s . The gain margin of the is system is (A) 11.95 dB (B) 17.67 dB (C) 21.33 dB (D) 23.9 dB

Common Data for Question 39 and 40 : A system is described by the following state and output equations dx1 (t) =- 3x1 (t) + x2 (t) + 2u (t) dt dx2 (t) =- 2x2 (t) + u (t) dt y (t) = x1 (t) when u (t) is the input and y (t) is the output

EE SP 5.39

EE SP 5.40

nodia The system transfer function is (A) 2 s + 2 s + 5s - 6 (C) 2 2s + 5 s + 5s + 6

s+3 s + 5s + 6 (D) 2 2s - 5 s + 5s - 6 (B)

2

The state-transition matrix of the above system is e - 3t e - 2t - e - 3t 0 e - 3t (B) (A) = - 2t G G = 0 e - 2t e + e - 3t e - 2t e - 3t e - 2t + e - 3t (C) = G 0 e - 2t

e3t e - 2t - e - 3t (D) = G 0 e - 2t


ONE MARK

A function y (t) satisfies the following differential equation : dy (t) + y (t) = d (t) dt where d (t) is the delta function. Assuming zero initial condition, and denoting the unit step function by u (t), y (t) can be of the form (A) et (B) e - t (C) et u (t)

(D) e - t u (t)


TWO MARK

The transfer function of a linear time invariant system is given as G (s) = 2 1 s + 3s + 2 The steady state value of the output of the system for a unit impulse input applied at time instant t = 1 will be (A) 0 (B) 0.5 (C) 1 (D) 2


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Control Systems

Page 289

The transfer functions of two compensators are given below : 10 (s + 1) C1 = , C2 = s + 10 (s + 10) 10 (s + 1) Which one of the following statements is correct ? (A) C1 is lead compensator and C2 is a lag compensator (B) C1 is a lag compensator and C2 is a lead compensator (C) Both C1 and C2 are lead compensator (D) Both C1 and C2 are lag compensator

EE SP 5.44

The asymptotic Bode magnitude plot of a minimum phase transfer function is shown in the figure :

nodia This transfer function has (A) Three poles and one zero (B) Two poles and one zero (C) Two poles and two zero (D) One pole and two zeros EE SP 5.45

Figure shows a feedback system where K > 0

The range of K for which the system is stable will be given by (A) 0 < K < 30 (B) 0 < K < 39 (C) 0 < K < 390 (D) K > 390 EE SP 5.46

The transfer function of a system is given as 100 s2 + 20s + 100 The system is (A) An over damped system (B) An under damped system (C) A critically damped system (D) An unstable system


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Control Systems

Chapter 5

Statement for Linked Answer Question 47 and 48 . The state space equation of a system is described by Xo = AX + Bu,Y = CX where X is state vector, u is input, Y is output and 0 1 0 A == , B = = G, C = [1 0] G 0 -2 1 EE SP 5.47

The transfer function G(s) of this system will be s (B) s + 1 (A) s (s - 2) (s + 2) s 1 (C) (D) (s - 2) s (s + 2)

EE SP 5.48

A unity feedback is provided to the above system G (s) to make it a closed loop system as shown in figure.

nodia

For a unit step input r (t), the steady state error in the input will be (A) 0 (B) 1 (C) 2 (D) 3 YEAR 2007

EE SP 5.49

ONE MARK

The system shown in the figure is

(A) Stable (B) Unstable (C) Conditionally stable (D) Stable for input u1 , but unstable for input u2 YEAR 2007 EE SP 5.50

TWO MARKS

If x = Re [G (jw)], and y = Im [G (jw)] then for w " 0+ , the Nyquist plot for G (s) = 1/s (s + 1) (s + 2) is (A) x = 0 (B) x =- 3/4 (C) x = y - 1/6 (D) x = y/ 3


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Control Systems

Page 291

The system 900/s (s + 1) (s + 9) is to be such that its gain-crossover frequency becomes same as its uncompensated phase crossover frequency and provides a 45c phase margin. To achieve this, one may use (A) a lag compensator that provides an attenuation of 20 dB and a phase lag of 45c at the frequency of 3 3 rad/s (B) a lead compensator that provides an amplification of 20 dB and a phase lead of 45c at the frequency of 3 rad/s (C) a lag-lead compensator that provides an amplification of 20 dB and a phase lag of 45c at the frequency of 3 rad/s (D) a lag-lead compensator that provides an attenuation of 20 dB and phase lead of 45c at the frequency of 3 rad/s

EE SP 5.52

nodia

If the loop gain K of a negative feed back system having a loop transfer function K (s + 3) / (s + 8) 2 is to be adjusted to induce a sustained oscillation then (A) The frequency of this oscillation must be 4 3 rad/s

(B) The frequency of this oscillation must be 4 rad/s

(C) The frequency of this oscillation must be 4 or 4 3 rad/s (D) Such a K does not exist

EE SP 5.53

The system shown in figure below

can be reduced to the form

with (A) X = c0 s + c1, Y = 1/ (s2 + a0 s + a1), Z = b0 s + b1 (B) X = 1, Y = (c0 s + c1) / (s2 + a0 s + a1), Z = b0 s + b1 (C) X = c1 s + c0, Y = (b1 s + b0) / (s2 + a1 s + a0), Z = 1 (D) X = c1 s + c0, Y = 1/ (s2 + a1 s + a), Z = b1 s + b0 EE SP 5.54

Consider the feedback system shown below which is subjected to a unit step input. The system is stable and has following parameters Kp = 4, Ki = 10, w = 500 and


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Control Systems

Chapter 5

x = 0.7 .The steady state value of Z is

(A) 1

(B) 0.25

(C) 0.1

(D) 0


nodia R-L-C circuit shown in figure

EE SP 5.55

EE SP 5.56

For a step-input ei , the overshoot in the output e0 will be (A) 0, since the system is not under damped (B) 5 % (C) 16 % (D) 48 %

If the above step response is to be observed on a non-storage CRO, then it would be best have the ei as a (A) Step function (B) Square wave of 50 Hz (C) Square wave of 300 Hz (D) Square wave of 2.0 KHz YEAR 2006

ONE MARK

3 (s - 2) , 4s2 - 2s + 1 the matrix A in the state space form Xo = AX + Bu is equal to V V R R S0 1 0 W S1 0 0 W (B) S 0 0 1 W (A) S 0 1 0 W SS- 1 2 - 4 WW SS- 1 2 - 4 WW XV TR V X RT 0 1 1 0 0 0 W W S S (C) S3 - 2 1 W (D) S 0 0 1 W SS1 - 2 4 WW SS- 1 2 - 4 WW X in 4 Volumes by RK Kanodia X T T Shop GATE Electrical at maximum discount at

EE SP 5.57

For a system with the transfer function H (s) =

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Control Systems

Page 293


TWO MARKS

Consider the following Nyquist plots of loop transfer functions over w = 0 to w = 3 . Which of these plots represent a stable closed loop system ?

nodia (A) (1) only (C) all, except (3)

EE SP 5.59

The Bode magnitude plot H (jw) =

(B) all, except (1) (D) (1) and (2) only 10 4 (1 + jw) is (10 + jw) (100 + jw) 2


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EE SP 5.60

Control Systems

Chapter 5

A closed-loop system has the characteristic function (s2 - 4) (s + 1) + K (s - 1) = 0 . Its root locus plot against K is

nodia YEAR 2005

ONE MARK

EE SP 5.61

A system with zero initial conditions has the closed loop transfer function. s2 + 4 T (s) = (s + 1) (s + 4) The system output is zero at the frequency (A) 0.5 rad/sec (B) 1 rad/sec (C) 2 rad/sec (D) 4 rad/sec

EE SP 5.62

Figure shows the root locus plot (location of poles not given) of a third order system whose open loop transfer function is


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Control Systems

(A) K3 s (C)

EE SP 5.63

Page 295

K s (s + 1) K (D) 2 s (s - 1) (B)

K s (s + 1) 2

2

nodia

The gain margin of a unity feed back control system with the open loop transfer (s + 1) function G (s) = is s2 (A) 0 (B) 1 2 (C) 2 (D) 3 YEAR 2005

EE SP 5.64

TWO MARKS

A unity feedback system, having an open loop gain K (1 - s) , G (s) H (s) = (1 + s)

becomes stable when (A) K > 1 (C) K < 1 EE SP 5.65

(B) K > 1 (D) K < - 1

When subject to a unit step input, the closed loop control system shown in the figure will have a steady state error of

(A) - 1.0 (C) 0 EE SP 5.66

(B) - 0.5 (D) 0.5

In the G (s) H (s)-plane, the Nyquist plot of the loop transfer function G (s) H (s) = pes passes through the negative real axis at the point (A) (- 0.25, j0) (B) (- 0.5, j0) (C) 0 (D) 0.5 - 0.25s

EE SP 5.67

If the compensated system shown in the figure has a phase margin of 60c at the


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Control Systems

Chapter 5

crossover frequency of 1 rad/sec, then value of the gain K is

(A) 0.366

(B) 0.732

(C) 1.366

(D) 2.738

Common Data For Q. 68 and 69 0 1 1 A state variable system Xo (t) = = X (t) + = Gu (t) with the initial condition G 0 -3 0 T X (0) = [- 1, 3] and the unit step input u (t) has

EE SP 5.68

nodia The state transition matrix - 3t 1 1 ) 3 (1 - e (A) = G - 3t 0 e 1 (C) > 0

EE SP 5.69

1 3

(e3 - t - e- 3t) H e- 3t

(e- t - e- 3t) H e- t

1 (B) > 0

1 3

1 (D) > 0

(1 - e- t) H e- t

The state transition equation t - e-t (A) X (t) = = - t G e

1 - e-t (B) X (t) = = - 3t G 3e

t - e 3t (C) X (t) = = - 3t G 3e

t - e - 3t (D) X (t) = = - t G e

YEAR 2004

ONE MARK

EE SP 5.70

The Nyquist plot of loop transfer function G (s) H (s) of a closed loop control system passes through the point (- 1, j 0) in the G (s) H (s)plane. The phase margin of the system is (A) 0c (B) 45c (C) 90c (D) 180c

EE SP 5.71

Consider the function, F (s) =

5 s (s + 3s + 2) where F (s) is the Laplace transform of the of the function f (t). The initial value of f (t) is equal to (A) 5 (B) 25

(C) EE SP 5.72

5 3

2

(D) 0

For a tachometer, if q (t) is the rotor displacement in radians, e (t) is the output voltage and Kt is the tachometer constant in V/rad/sec, then the transfer E (s) will be function, Q (s) (A) Kt s2

(B) Kt s


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Control Systems

(C) Kt s

Page 297

(D) Kt


For the equation, s3 - 4s2 + s + 6 = 0 the number of roots in the left half of s -plane will be (A) Zero (B) One (C) Two

EE SP 5.74

EE SP 5.76

C (s) is equal to R (s)

nodia 2 (B) s + s2 + 1 s (D) 2 1 s +s+1

o = AX where The state variable description of a linear autonomous system is, X X is the two dimensional state vector and A is the system matrix given by 0 2 A = = G. The roots of the characteristic equation are 2 0 (A) - 2 and + 2 (B) - j2 and + j2 (C) - 2 and - 2 (D) + 2 and + 2 The block diagram of a closed loop control system is given by figure. The values of K and P such that the system has a damping ratio of 0.7 and an undamped natural frequency wn of 5 rad/sec, are respectively equal to

(A) 20 and 0.3 (C) 25 and 0.3 EE SP 5.77

(D) Three

For the block diagram shown, the transfer function

2 (A) s +2 1 s 2 (C) s + s + 1 s

EE SP 5.75

TWO MARKS

(B) 20 and 0.2 (D) 25 and 0.2

The unit impulse response of a second order under-damped system starting from rest is given by c (t) = 12.5e - 6t sin 8t, t $ 0 . The steady-state value of the unit step response of the system is equal to (A) 0 (B) 0.25 (C) 0.5 (D) 1.0


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Control Systems

In the system shown in figure, the input x (t) = sin t . In the steady-state, the response y (t) will be

1 sin (t - 45c) 2 (C) sin (t - 45c) (A)

EE SP 5.79

Chapter 5

1 sin (t + 45c) 2 (D) sin (t + 45c)

(B)

The open loop transfer function of a unity feedback control system is given as 1. G (s) = as + 2 s The value of ‘a ’ to give a phase margin of 45c is equal to (B) 0.441 (A) 0.141 (C) 0.841 (D) 1.141

nodia YEAR 2003

EE SP 5.80

EE SP 5.81

ONE MARK

A control system is defined by the following mathematical relationship d2 x + 6 dx + 5x = 12 (1 - e - 2t) dt dt2 The response of the system as t " 3 is (A) x = 6 (B) x = 2 (C) x = 2.4 (D) x =- 2 A lead compensator used for a closed loop controller has the following transfer K (1 + as ) function For such a lead compensator (1 + bs ) (A) a < b (B) b < a (C) a > Kb (D) a < Kb

EE SP 5.82

2 A second order system starts with an initial condition of = G without any external 3 e - 2t 0 input. The state transition matrix for the system is given by = G. The state 0 e-t of the system at the end of 1 second is given by 0.271 (A) = 1.100G

0.135 (B) = 0.368G

0.271 (C) = 0.736G

0.135 (D) = 1.100 G


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Control Systems


Page 299

TWO MARKS

A control system with certain excitation is governed by the following mathematical equation d2 x + 1 dx + 1 x = 10 + 5e- 4t + 2e- 5t 2 dt 18 dt2 The natural time constant of the response of the system are (A) 2 sec and 5 sec (B) 3 sec and 6 sec (C) 4 sec and 5 sec (D) 1/3 sec and 1/6 sec

EE SP 5.84

The block diagram shown in figure gives a unity feedback closed loop control system. The steady state error in the response of the above system to unit step input is

nodia (A) 25% (B) 0.75 % (C) 6% (D) 33%

EE SP 5.85

The roots of the closed loop characteristic equation of the system shown above (Q-5.55)

(A) - 1 and - 15 (B) 6 and 10 (C) - 4 and - 15 (D)- 6 and - 10 EE SP 5.86

The following equation defines a separately excited dc motor in the form of a differential equation d2 w + B dw + K2 w = K V dt J dt LJ LJ a The above equation may be organized in the state-space form as follows R 2 V Sd w W dw S dt2 W = P dt + QV > H a S dw W w S dt W X T


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Control Systems

Where the P matrix is given by K - B - LJ (A) = J G 1 0 0 1 (C) =- K - B G LJ J 2

2

EE SP 5.87

Chapter 5

- K - BJ (B) = LJ G 0 1 1 0 (D) =- B - K G J LJ 2

2

The loop gain GH of a closed loop system is given by the following expression K The value of K for which the system just becomes unstable is s (s + 2) (s + 4) (A) K = 6 (B) K = 8 (C) K = 48 (D) K = 96

EE SP 5.88

nodia

The asymptotic Bode plot of the transfer function K/ [1 + (s/a)] is given in figure. The error in phase angle and dB gain at a frequency of w = 0.5a are respectively

(A) 4.9c, 0.97 dB (C) 4.9c, 3 dB EE SP 5.89

(B) 5.7c, 3 dB (D) 5.7c, 0.97 dB

The block diagram of a control system is shown in figure. The transfer function G (s) = Y (s) /U (s) of the system is

1 18^1 + 12s h^1 + s3 h 1 (C) 27^1 + 12s h^1 + s9 h (A)

1 27^1 + s6 h^1 + s9 h 1 (D) 27^1 + s9 h^1 + s3 h

(B)

***********


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Control Systems

Page 301

SOLUTION SOL 5.1

Correct option (B). In the given Routh-Hurwitz array of polynomial, all the elements of a row have zero value. This is due to symmetrical location of the roots in the s -plane with respect to origin. The system is either marginally stable or unstable. Now, we check this characteristic for all the given options. Option (A): Only one root is at origin. So, it does not satisfy the symmetrical condition. Option (B): Since, the system has imaginary roots, so we get the pole-zero location diagram as shown below.

nodia

The imaginary roots on jw (imaginary) axis are symmetrical with respect to origin. Hence, this option is correct. Option (C): The system has only positive real roots as shown below. So, the root location diagram does not satisfy the symmetrical condition.

Option (D): Again, the system has only negative real roots, as shown below. So, the root location diagram does not satisfy the symmetrical condition.

SOL 5.2

Correct option (C). We have the root locus diagram as


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Control Systems

Chapter 5

As the root locus have poles s =- 1, - 2 and root lies in even multiple of poles, so it is converse of the main transfer function. Hence, gain should be negative, i.e. -K G ^s h H ^s h = ^s + 1h^s + 2h This is open loop transfer function and closed loop transfer function is given by G ^s h H ^s h C ^s h = 1 + G ^s h H ^s h R ^s h -K ^s + 1h^s + 2h = -K 1+ ^s + 1h^s + 2h -K = ^s + 1h^s + 2h - K SOL 5.3

nodia Correct answer is 0.618. The Block diagram of given system is

The open loop transfer function is

^s + ah s + ^1 + ah s + â - 1h s + ^1 - ah So, we obtain the character equation as 1 + G ^s h H ^s h = 0 G ^s h H ^s h =

or or or or For

3

2

^s + ah =0 s + ^1 + ah s + â - 1h s + ^1 - ah s3 + ^1 + ah s2 + â - 1h s + ^1 - ah + ^s + ah = 0 s3 + ^1 + ah s2 + â - 1 + 1h s + 1 - a + a = 0 s3 + ^1 + ah s2 + as + 1 = 0 the characteristic equation, we form the Routh’s array as 1 a s3 2 1+a 1 s 1 1 + a a ^ h 0 s1 1+a 0 1 s 1+

3

2

For stable system, the required condition is or or

1+a > 0 a >-1 a ^1 + ah - 1 >0 1+a

or a ^1 + ah - 1 > 0 Solving the inequality, we obtain the roots a = -1 2

5 , -1 + 2

5


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Control Systems

Page 303

So, we get the result for inequality as a > 0.618 and a < - 1.62 i.e. the minimum value of a is a = 0.618 SOL 5.4

Correct answer is 0.75. Given the Bode magnitude plot of the transfer function,

nodia Also, from the given transfer function, we have K ^1 + 0.55h^1 + as h G ^s h = s a1 + s k^1 + bs ha1 + s k 8 36

K ^1 - s/2h^1 + s/ a1 h = s a1 + s kb1 + s la1 + s k 8 36 1/b The first slope - 6 dB/octave is due to one pole that is 1/s Then, slope 0 dB/octave is due to addition of a zero in T.F. ^1 + s/2h. Again, + 6 dB/octave slope is due to one zero at corner frequency wC = 4 . Comparing it to the transfer function, we get ^1 + as h = ^1 + s/4h or a = 1/4 Similarly, at wC = 24 , there is an addition of a pole ^- 6 dB/octaveh. So, we get ^1 + bs h = ^1 + s/24h or b = 1 24 From the shown Bode plot, we observe that if we extended the slope - 6 dB/octave , it meets the frequency axis at wC = 8 . So, we have 0 = 20 log KX s w =8 or 1=K 8 C

or K =8 Therefore, we obtain the desired value as a = 1/4 = 24 = 0.75 1 4#8 bK 24 # 8


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Control Systems

Chapter 5

Correct answer is 0. Closed loop T.F. T ^s h = G ^s h 1 + G ^s h 1 + G ^s h G ^s h 1+ 1 G ^s h 1 G ^s h

=

4 s2 + 0.4s + 4 4 s2 + 0.4s + 4

2 = s + 0.4s + 4 4 2 = s + 0.4s + 4 4

2 = s + 0.4s + 4 - 4 4 4 G ^s h = s ^s + 0.4h K p = lim G ^s h s"0

Open loop. T.F.

nodia Error constant,

4 =3 s ^s + 0.4h ess = 1 = 0 1 + Kp = lim s"0

Steady state error

SOL 5.6


1 0 x1 1 xo1 > o H = >1 1H>x H + >1H U x2 2 Y = AX + BU

State transmission matrix

f ^ t h = L-1 ^sI - Ah-1 s-1 0 6SI - A@ = >1 s - 1H 1 >s - 1 0 H ^s - 1h2 1 s - 1 1 0 s-1 = L-1 > 1 1 H

f ^ t h = L-1

So,

^s - 1h2

s-1

t

e 0 f ^ t h = > t tH te e SOL 5.7

Correct answer is 5. G ^s h =

K s ^s + 2h^s2 + 2s + 2h G ^s h Closed loop T.F. = 1 + G ^s h K s ^s + 2h^s2 + 2s + 2h = K 1+ s ^s + 2h^s2 + 2s + 2h Closed loop T.F. =

K s ^s + 2h^s2 + 2s + 2h + K

Characteristics equation


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Control Systems

Page 305

s ^s + 2h^s2 + 2s + 2h + K = 0 ^s2 + 2s h^s2 + 2s + 2h + K = 0 s 4 + 4s3 + 4s2 + 4s + K = 0 Routh array s4

1

6

k

3

4

4

0

s2

5

k

0

s1

20 - 4k 5

0

s0

k

s

nodia For marginally stable

20 - 4k = 0

k =5

or

SOL 5.8


5 ^s + 4h ^s + 0.25h^s2 + 4s + 25h ^1 + s/4h = 5#4 0.25 # 25 s 4s s2 a1 + 0.25 kb1 + 25 + 25 l ^1 + s/4h = 3.2 # s 4s s2 a1 + 0.25 kb1 + 25 + 25 l

G ^s h =

So constant gain term, Highest corner frequency SOL 5.9

K = 3.2 wH = 5

Correct option is (C). For controllability, we check controllability matrix CM = 6B AB@

or,

So,

CM = 6Q PQ@ -1 1 0 1 H> H = > H PQ = > 0 -3 1 -3 0 1 H CM = > 1 -3

det ^CM h =- 1 ! 0 So, the system is controllable. For observalibility, we check observalibility matrix C OM = > H CA or

R OM = > H RP -1 1 H = 80 - 3B RP = 80 1B> 0 -3


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Control Systems

So,

Chapter 5

0 1 OM = > 0 - 3H

det ÔM h = 0 So, the system is not observable. SOL 5.10

Correct option is (C). We have the signal flow graph of the system as

nodia

Now, we have to obtain the transfer function, C ^s h G ^s h = U ^s h From the SFG, we have two forward paths given as p1 = h 0 1 1 1 1 1 = h20 s s s p 2 = h1 1 1 = h1 s s The graph determinant is given by T = 1 - (sum of all loop gains) + (sum of all gain products of 2 non-touching loops) - (sum of all gain products of 3 non-touching loops) + ... T = 1 - a - a1 + - a2 0 k s s = 1 + a1 + a20 s s Also, we have the determinants for the forward paths p1 and p2 as T1 = 1 T 2 = 1 - a - a1 k = 1 + a1 s s Thus, using Mason’s gain formula, we obtain the transfer function as pk D k p D + p2 D 2 T.F. = = 1 1 D D h 0 1 + h1 1 + a1 2^ h sa sk = s 1 + a1 + a20 s s h 0 + h1 ^s + a1h = 2 s + a1 s + a 0 Since, we have So,

/


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Control Systems

Page 307

h1 = b 1 and h 0 = b 0 - b 1 a1 Substituting it in equation (1), we get T.F. = or

SOL 5.11

G ^s h =

b 0 + b1 s s + a1 s + a 0 2

C ^s h = 2 b 0 + b1 s V ^s h s + a1 s + a 0

Correct option is (A). For the given system, we have

G ^s h H ^s h < 1 So, we may easily conclude that the Nyquist-plot intersect the negative real axis between 0 and - 1 point, i.e. the Nyquist plot does not enclose the point ^- 1, 0h or in other words number of encirclements is zero.

nodia N =0 From Nyquist criterion, we have the relation

...(1)

...(2) N = P-Z where P is the number of open loop poles in right-half of s -plane and Z is the number of closed loop poles (roots of characteristic equation) in right-half of s -plane. Since, for a stable closed loop system, we must have Z =0 Thus, from equations (1), (2), and (3), we get

...(3)

P =0 i.e. no poles of open loop system lies in right-half of s -plane, or in other words, all the poles lies in left-half of s -plane. Hence, the given system is stable if all poles of G ^s h H ^s h are in left half of the s -plane. SOL 5.12

Correct option is (B). We have the block diagram of given system as

By using minimization technique, we get

The closed loop transfer function for the dotted box is


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Control Systems

Chapter 5

G ^s h 1 + G ^s h

Again, we obtain the closed loop transfer function for the dotted box as sG ^s h ^s + 1h G ^s h + 1 So, the block diagram reduces to

nodia Thus, we have the overall transfer function as G ^s h ^s + 1h G ^s h + 1 T.F. 2 ^s + s h G ^s h + s + G ^s h s 6^s + 1h G ^s h + 1@ sG ^s h = 2 ^s + s + 1h G ^s h + s ALTERNATIVE METHOD :

For the given block diagram, we draw the signal flow graph as

From the signal flow graph, we have forward paths p1 = 1 G ^ s h s = G ^ s h s and the closed loops G ^s h ; L11 = 1 G ^s h^- 1h =s s L12 = G ^s h^- 1h =- G ^s h;


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Control Systems

Page 309

L13 = G ^s h s ^- 1h

and

=- sG ^s h So, the transfer function is obtained as pT T.F. = 1 1 T =

G ^s h^1 h G ^s h 1 - ;- G ^s h - sG ^s hE s

sG ^s h s + G ^s h^s2 + s + 1h For any value of G ^s h given in the options, the desired system transfer function can not be achieved. T.F. =

SOL 5.13

nodia Correct option is (A). We have the state space equation, 0 1 x1 0 xo1 > H => H> H + > H u o 1 - 1 - 1 x2 x2

x1 y = 81 0B> H x2 From the state-space equation, we obtain the matrices 0 1 H A => -1 -1 0 B => H 1

C = 81 0B So, we obtain S 0

0

1

S

-1

H-> H=> H 6SI - A@ = > 0 S -1 -1 1 S+1

Therefore, S+1 1 1 > H S2 + S + 1 - 1 S Hence, we obtain the transfer function of the system as T.F. = C [SI - A] -1 B S+1 1 0 > H> H = 81 0B 2 1 S + S + 1 -1 S 1 1 81 0B> H = 2 1 S S +S+1 = 2 1 S +S+1 This is the closed loop transfer function, So we get G (S ) = 2 1 1 + G (S ) S +S+1 G (S ) = 2 1 S +S Thus, the steady state error for unit step input is

6SI - A@-1 =


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Control Systems

Chapter 5

1 1 + lim G (S ) S"0 1 = 1 + lim 2 1 d"0 d + d = 1 =0 1+3

eSS =

SOL 5.14

Correct option is (C). Given the magnitude Bode plot,

nodia

Since, slope at 1/3 is increasing and slope at 1 is decreasing. So, we can write frequency response function as 1 b 3 + jw l G ^ jwh = ^1 + jwh jw 1 3 c1 + 1/3 m or G ^ jwh = ^1 + jwh a ^1 + jw/a h or G ^ jwh = b ^1 + jw/b h Therefore, the maximum phase angle is f m = 90c - 2 tan-1 _ a/b i

= 90c - 2 tan-1 ^1/ 3 h = 90c - 60c = 30c and the maximum phase occur at w = ab = 1/ 3 Thus, we obtain the gain at the frequency as 2 2 2 1/2 Gm = 20 log c 1 m + b 1 l - 20 log e 12 + c 1 m o 3 3 3 = 4.76 SOL 5.15

Correct option is (B). The Laplace transform of unit step fun n is U ^s h = 1 s So, the O/P of the system is given as Y ^s h = b 1 lb 1 l = 12 s s s For zero initial condition, we check


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Control Systems

u^t h =

dy ^ t h dt

U ^s h = SY ^s h - y ^0 h U ^s h = s c 12 m - y ^0 h s 1 U ^s h = s

& & or,

Page 311

^y ^0 h = 0h

Hence, the O/P is correct which is Y ^s h = 12 s its inverse Laplace transform is given by y ^ t h = tu ^ t h

SOL 5.16

nodia Correct option is (B). From the given plot, we obtain the slope as Slope =

20 log G2 - 20 log G1 log w2 - log w1

From the figure

20 log G2 =- 8 dB 20 log G1 = 32 dB w1 = 1 rad/s w2 = 10 rad/s

and

So, the slope is Slope = - 8 - 32 log 10 - log 1 =- 40 dB/decade Therefore, the transfer function can be given as G ^s h = k2 S At w = 1, G ^ jwh = k 2 = k w In decibel,

20 log G ^ jwh = 20 log k = 32 32

or, k = 10 = 39.8 Hence, the Transfer function is G ^s h = k2 = 392.8 s s SOL 5.17

20

Correct option is (A). For the given SFG, we have two forward paths Pk1 = ^1 h^s-1h^s-1h^1 h = s-2 Pk2 = ^1 h^s-1h^1 h^1 h = s-1 since, all the loops are touching to both the paths Pk1 and Pk2 so, D k 1 = Dk 2 = 1 Now, we have D = 1 - (sum of individual loops)


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Control Systems

Chapter 5

+ (sum of product of nontouching loops) Here, the loops are

L1 = ^- 4h^1 h =- 4 L2 = ^- 4h^s-1h = 4s-1 L 3 = ^- 2h^s-1h^s-1h =- 2s-2

L 4 = ^- 2h^s-1h^1 h =- 2s-1 As all the loop L1, L2, L 3 and L 4 are touching to each other so, D = 1 - ^L1 + L2 + L 3 + L 4h = 1 - ^- 4 - 4s-1 - 2s-2 - 2s-1h

= 5 + 6s-1 + 2s-2 From Mason’s gain formulae Y ^s h s-2 + s-1 s+1 = SPk Dk = 2 -1 -2 = D U ^s h 5s + 6s + 2 5 + 6s + 2s

SOL 5.18

nodia Correct option is (C). Given, open loop transfer function

G ^s h = 10Ka = Ka 1 1 + 10s s + 10 By taking inverse Laplace transform, we have g ^ t h = e- t 1 10

Comparing with standard form of transfer function, Ae-t/t , we get the open loop time constant, tol = 10 Now, we obtain the closed loop transfer function for the given system as G ^s h 10Ka = H ^s h = 1 + G ^s h 1 + 10s + 10Ka Ka = s + ^Ka + 101 h By taking inverse laplace transform, we get h ^ t h = ka .e-^k + ht So, the time constant of closed loop system is obtained as tcl = 1 1 ka + 10 or, tcl = 1 ka a

1 10

(approximately)

Now, given that ka reduces open loop time constant by a factor of 100. i.e., tcl = tol 100 1 = 10 or, 100 ka or, SOL 5.19

ka = 10

Correct option is (A). Given, the state variable formulation, - 2 0 x1 1 xo1 > o H = > 0 - 1H>x H + >1H u x2 2

....(1)


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Control Systems

Page 313

x1 y = 61 0@> H x2

and

....(2)

From Eq. (1) we get xo1 = 2x1 + u Taking Laplace transform sX1 - x1 ^0 h =- 2X1 + 1 (Here, X1 denotes Laplace transform of x1 ) s So, ^s + 2h X1 = s1 ^x1 ^0 h = 0h 1 or, X1 = s ^s + 2h Now, from Eq. (2) we have

nodia

....(3)

y = x1 Taking Laplace transform both the sides, Y = XL

or,

or,

1 s ^s + 2h Y = 1 ;1 - 1 E 2 s s+2 Y =

(from eq. (3))

Taking inverse Laplace transform y = 1 8u ^ t h - e-2t u ^ t hB 2 = 1 - 1 e-2t 2 2

SOL 5.20

^for t > 0h

Correct option is (A). From the given state variable system, we have -2 0 A => 0 1H and

1 B = > H; C = 61 0@ 1

Now, we obtain the controllability matrix CM = 6B : AB@ 1 -2 => 2 1H

and the observability matrix is obtained as C 1 0 H OM = > H = > CA -2 0 So, we get

Rank of the controllability matrix " Rank ^CM h = 2 Rank of the observability matrix " Rank ÔM h = 1 Since, the order of state variable is 2 ^x1 and x2h. Therefore, we have Rank ^CM h = order of state variables but, Rank (OM ) < order of state variables Thus, system is controllable but not observable


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Control Systems

Chapter 5

Correct option is (D). General form of state equations are given as xo = Ax + Bu yo = Cx + Du For the given problem R 0 a 0V R0V 1 S W S W A = S 0 0 a2W, B = S0W SSa SS1WW 0 0WW 3 TR XVR V R VT X 0 a 0 1 S WS0W S 0W AB = S 0 0 a2WS0W = Sa2W SSa 0 0WWSS1WW SS 0WW 3 TR XT X VTR XV R 0 0 a1 a2WS0W Sa1 a2VW S A2 B = Sa2 a 3 0 0WS0W = S 0W SS 0 a a 0WWSS1WW SS 0WW 3 1 T XT X T X For controllability it is necessary that following matrix has a tank of n = 3 . R0 0 a a V 1 2W S 2 U = 6B : AB : A B@ = S0 a2 0W SS1 0 0WW So, a2 ! 0 T X a1 a 2 ! 0 & a1 ! 0 a 3 may be zero or not.

SOL 5.22

nodia Correct option is (A).

Y (s) =

K (s + 1) [R (s) - Y (s)] s + as2 + 2s + 1 3

K (s + 1) s + as2 + 2s + 1E K (s + 1) R (s) = 3 s + as2 + 2s + 1 Y (s) [s3 + as2 + s (2 + k) + (1 + k)] = K (s + 1) R (s) Transfer Function, Y (s) K (s + 1) H (s) = = R (s) s3 + as2 + s (2 + k) + (1 + k) Routh Table : Y (s) ;1 +

3

For oscillation, a (2 + K) - (1 + K) =0 a a = K+1 K+2 Auxiliary equation


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Control Systems

Page 315

A (s) = as2 + (k + 1) = 0 s2 =- k + 1 a s2 = - k + 1 (k + 2) =- (k + 2) (k + 1) s = j k+2 jw = j k + 2 w =

(Oscillation frequency)

k =2 a = 2 + 1 = 3 = 0.75 2+2 4

and SOL 5.23

k+2 = 2


nodia

jw + a GC (s) = s + a = s+b jw + b Phase lead angle, f = tan-1 a w k - tan-1 a w k a b Jw - wN -1 K a b O = tan-1 w (b - a) = tan c ab + w 2 m 2 O KK 1+ w O ab P L For phase-lead compensation f > 0

b-a > 0 b >a Note: For phase lead compensator zero is nearer to the origin as compared to pole, so option (C) can not be true.

SOL 5.24

Correct option is (A). f = tan-1 a w k - tan-1 a w k a b 1/a 1/b df = =0 2 2 w dw 1 +a k 1 +awk a b 2 2 1 + w = 1+1w a ab2 b b a2 1 - 1 = w2 1 - 1 a b ab b a b l w = ab = 1#2 =

SOL 5.25

2 rad/ sec

Correct option is (A). Gain margin is simply equal to the gain at phase cross over frequency ( wp ). Phase cross over frequency is the frequency at which phase angle is equal to - 180c. From the table we can see that +G (jwp) =- 180c, at which gain is 0.5. 1 GM = 20 log 10 e G (jwp) o = 20 log b 1 l = 6 dB 0.5 Phase Margin is equal to 180c plus the phase angle fg at the gain cross over


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Control Systems

Chapter 5

frequency ( wg ). Gain cross over frequency is the frequency at which gain is unity. From the table it is clear that G (jwg) = 1, at which phase angle is - 150c fPM = 180c + +G (jwg) = 180 - 150 = 30c SOL 5.26

Correct option is (A). We know that steady state error is given by sR (s) ess = lim s " 0 1 + G (s) where

R (s) " input

G (s) " open loop transfer function For unit step input R (s) = 1 s sb 1 l s So = 0.1 ess = lim s " 0 1 + G (s)

nodia 1 + G (0) = 10

Given input or

G (0) = 9 r (t) = 10 [m (t) - m (t - 1)] R (s) = 10 :1 - 1 e-sD s s -s = 10 :1 - e D s

So steady state error (1 - e-s) s # 10 s el ss = lim s"0 1 + G (s) 10 (1 - e0) = =0 1+9 SOL 5.27

Correct option is (B). Transfer function having at least one zero or pole in RHS of s -plane is called non-minimum phase transfer function. s-1 G (s) = (s + 2) (s + 3) • In the given transfer function one zero is located at s = 1 (RHS), so this is a non-minimum phase system. • Poles - 2, - 3 , are in left side of the complex plane, So the system is stable

SOL 5.28

Correct option is (A). K bs + 2 l 3 G (s) = 2 s (s + 2) Steps for plotting the root-locus (1) Root loci starts at s = 0, s = 0 and s =- 2 (2) n > m , therefore, number of branches of root locus b = 3 (3) Angle of asymptotes is given by


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Control Systems

Page 317

(2q + 1) 180c , q = 0, 1 n-m (2 # 0 + 1) 180c (1) = 90c (3 - 1) (2 # 1 + 1) 180c (2) = 270c (3 - 1) (4) The two asymptotes intersect on real axis at centroid / Poles - / Zeroes x = n-m - 2 - b- 2 l 3 = =- 2 3-1 3 (5) Between two open-loop poles s = 0 and s =- 2 there exist a break away point. s2 (s + 2) K =2 bs + 3 l dK = 0 ds

nodia s =0 Root locus is shown in the figure

Three roots with nearly equal parts exist on the left half of s -plane. SOL 5.29

Correct option is (A). The system may be reduced as shown below

1 s (s + 1 + K ) Y (s) = 1 R (s) 1 + s (s + 1 + K ) 1 = 2 s + s (1 + K ) + 1 This is a second order system transfer function, characteristic equation is


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Control Systems

Chapter 5

s2 + s (1 + K) + 1 = 0 Comparing with standard form We get

s2 + 2xwn s + wn2 = 0 x = 1+K 2 M p = e- px/

Peak overshoot,

1 - x2

So the Peak overshoot is effected by K . SOL 5.30

Correct option is (C). Stability : Eigen value of the system are calculated as

nodia A - lI = 0 -1 2 l 0 A - lI = > -> H 0 2 0 lH -1 - l 2 => 0 2 - lH

A - lI = (- 1 - l) (2 - l) - 2 # 0 = 0 l1, l2 =- 1, 2 & Since eigen values of the system are of opposite signs, so it is unstable Controllability : -1 2 0 , B=> H A => H 0 2 1 2 AB = > H 2 0 2 [B: AB] = > H 1 2 Y 0 6B: AB@ = So it is controllable.

SOL 5.31

Correct option is (C). Given characteristic equation s (s + 1) (s + 3) + K (s + 2) = 0 ; s (s2 + 4s + 3) + K(s + 2) = 0 s3 + 4s2 + (3 + K) s + 2K = 0 From Routh’s tabulation method s3

1

3+K

s2

4

2K

s1 s0

4(3 + K) - 2K (1) 4

=

12 + 2K 4

K>0

>0

2K

There is no sign change in the first column of routh table, so no root is lying in


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Control Systems

Page 319

right half of s -plane. For plotting root locus, the equation can be written as K (s + 2) 1+ =0 s (s + 1) (s + 3) Open loop transfer function K (s + 2) G (s) = s (s + 1) (s + 3) Root locus is obtained in following steps: 1. No. of poles n = 3 , at s = 0, s =- 1 and s =- 3 2. No. of Zeroes m = 1, at s =- 2 3.

The root locus on real axis lies between s = 0 and s =- 1, between s =- 3 and s =- 2 .

4.

Breakaway point lies between open loop poles of the system. Here breakaway point lies in the range - 1 < Re [s] < 0 . Asymptotes meet on real axis at a point C , given by

nodia 5.

C =

/ real part of poles - / real parts of zeroes

n-m (0 - 1 - 3) - (- 2) = 3-1

=- 1 As no. of poles is 3, so two root loci branches terminates at infinity along asymptotes Re (s) =- 1

SOL 5.32

Correct option is (A). Given

1 s (s + 1) (s + 2) 1 G (jw) = jw (jw + 1) (jw + 2) 1 G (jw) = 2 w w + 1 w2+ 4 +G (jw) =- 90c - tan- 1 (w) - tan- 1 (w/2) G (s) =

In nyquist plot For w = 0, G (jw) = 3 +G (jw) =- 90c For w = 3, G (jw) = 0 +G (jw) =- 90c - 90c - 90c =- 270c Intersection at real axis 1 G (jw) = jw (jw + 1) (jw + 2) 1 = jw (- w2 + j3w + 2) - 3w2 - jw (2 - w2) 1 = # - 3w2 + jw (2 - w2) - 3w2 - jw (2 - w2) - 3w2 - jw (2 - w2) = 9w4 + w2 (2 - w2) 2 2 jw (2 - w2) = 4 -23w 2 2 4 9w + w (2 - w ) 9w + w2 (2 - w2) 2


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Control Systems

Chapter 5

Im [G (jw)] = 0 w (2 - w2) =0 9w4 + w2 (2 - w2) 2 - w2 = 0 & w = 2 rad/sec 2 rad/sec, magnitude response is 1 G (jw) at w = 2 = 2 2+1 2+4 =1<3 6 4

At real axis So, At w =

SOL 5.33

SOL 5.34

Correct option is (D). Overall gain of the system is written as G = G1 G 2 1 G3 We know that for a quantity that is product of two or more quantities total percentage error is some of the percentage error in each quantity. so error in overall gain G is 3 G = e1 + e2 + 1 e3

nodia Correct option is (D). Given Routh’s tabulation. s3

2

2

s2

4

4

s1

0

0

So the auxiliary equation is given by, 4s2 + 4 = 0 s2 =- 1 s =! j From table we have characteristic equation as 2s3 + 2s + 4s2 + 4 = 0 s3 + s + 2s2 + 2 = 0 s (s2 + 1) + 2 (s2 + 1) = 0 (s + 2) (s2 + 1) = 0 s =- 2 , s = ! j SOL 5.35

Correct option is (D). From Nyquist stability criteria, no. of closed loop poles in right half of s -plane is given as Z = P-N P " No. of open loop poles in right half s -plane N " No. of encirclement of (- 1, j0)


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Control Systems

Page 321

Here N =- 2 (` encirclement is in clockwise direction) P = 0 (` system is stable) So, Z = 0 - (- 2) Z = 2 , System is unstable with 2-poles on RH of s -plane. SOL 5.36

nodia

Correct option is (B). Since initial slope of the bode plot is - 40 dB/decade, so no. of poles at origin is 2. Transfer function can be written in following steps: 1. Slope changes from - 40 dB/dec. to - 60 dB/dec. at w1 = 2 rad/sec., so at w1 there is a pole in the transfer function. 2. Slope changes from - 60 dB/dec to - 40 dB/dec at w2 = 5 rad/sec., so at this frequency there is a zero lying in the system function. 3. The slope changes from - 40 dB/dec to - 60 dB/dec at w3 = 25 rad/sec, so there is a pole in the system at this frequency.

Transfer function K (s + 5) s (s + 2) (s + 25) Constant term can be obtained as. T (s) =

2

T (jw) at w = 0.1 = 80 So,

80 = 20 log

K (5) (0.1) 2 # 50

K = 1000 therefore, the transfer function is 1000 (s + 5) T (s) = 2 s (s + 2) (s + 25) SOL 5.37

Correct option is (D). From the figure we can see that steady state error for given system is ess = 1 - 0.75 = 0.25 Steady state error for unity feed back system is given by sR (s) ess = lim = G s " 0 1 + G (s) s ^ s1 h 1 = lim > H; R (s) = s (unit step input) K s"0 1 + (s + 1)(s + 2) 1 = = 2 2+K 1 + K2 So, ess = 2 = 0.25 2+K


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Control Systems

Chapter 5

2 = 0.5 + 0.25K K = 1.5 = 6 0.25 SOL 5.38

Correct option is (D). Open loop transfer function of the figure is given by, - 0.1s G (s) = e s G (jw) = e

- j0.1w

jw Phase cross over frequency can be calculated as, +G (jwp) =- 180c 180 b- 0.1wp # p l - 90c =- 180c 0.1wp # 180c = 90c p 0.1wp = 90c # p 180c

nodia wp = 15.7 rad/sec

So the gain margin (dB)

1 G (jwp) o = 20 log = 11 G ^ 15.7 h = 20 log 15.7 = 23.9 dB

= 20 log e

SOL 5.39

Correct option is (C). Given system equations dx1 (t) =- 3x1 (t) + x2 (t) + 2u (t) dt dx2 (t) =- 2x2 (t) + u (t) dt y (t) = x1 (t) Taking Laplace transform on both sides of equations. sX1 (s) =- 3X1 (s) + X2 (s) + 2U (s) (s + 3) X1 (s) = X2 (s) + 2U (s) Similarly sX2 (s) =- 2X2 (s) + U (s) (s + 2) X2 (s) = U (s) From equation (1) & (2) U (s) + 2U (s) (s + 3) X1 (s) = s+2 U (s) 1 + 2 (s + 2) X1 (s) = E s + 3; s + 2 (2s + 5) = U (s) (s + 2) (s + 3) From output equation,

...(1) ...(2)

Y (s) = X1 (s)


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Control Systems

So,

Y (s) = U (s)

Page 323

(2s + 5) (s + 2) (s + 3)

System transfer function Y (s) (2s + 5) T.F = = U (s) (s + 2) (s + 3) (2s + 5) = 2 s + 5s + 6 SOL 5.40

Correct option is (B). Given state equations in matrix form can be written as, - 3 1 x1 2 xo1 > o H = > 0 - 2H>x H + >1H u (t) x2 2 dX (t) = AX (t) + Bu (t) dt State transition matrix is given by

nodia f (t) = L- 1 6F (s)@ F (s) = (sI - A) - 1 s 0 -3 1 (sI - A) = > H - > 0 s 0 - 2H s + 3 -1 (sI - A) = > 0 s + 2H

(sI - A) - 1 =

So

SOL 5.41

s+2 1 1 > 0 s + 3H (s + 3) (s + 2)

V R 1 W S 1 (s + 3) (s + 3) (s + 2)W S -1 F (s) = (sI - A) = S W 1 S 0 (s + 2) W X T e- 3t e- 2t - e- 3t -1 f (t) = L [F (s)] = > H 0 e- 2t

Correct option is (D). Given differential equation for the function dy (t) + y (t) = d (t) dt Taking Laplace on both the sides we have, sY (s) + Y (s) = 1 (s + 1) Y (s) = 1 Y (s) =

1 s+1

Taking inverse Laplace of Y (s) y (t) = e- t u (t), t > 0 SOL 5.42

Correct option is (A). Given transfer function 1 s + 3s + 2 r (t) = d (t - 1)

G (s) = Input

2


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Control Systems

Chapter 5

R (s) = L [d (t - 1)] = e- s Output is given by Y (s) = R (s) G (s) -s = 2 e s + 3s + 2 Steady state value of output lim y (t) = lim sY (s)

t"3

s"0

se- s s " 0 s + 3s + 2 =0

= lim

SOL 5.43

2

Correct option is (A). For C1 Phase is given by

nodia qC = tan- 1 (w) - tan- 1 a w k 10 Jw - w N -1 K 10 O = tan 2 O KK w 1+ O 10 P L = tan- 1 c 9w 2 m > 0 (Phase lead) 10 + w 1

Similarly for C2 , phase is

qC2 = tan- 1 a w k - tan- 1 (w) 10 J w - wN - 1 K 10 O = tan 2 O KK w 1+ O 10 P L = tan- 1 c - 9w 2 m < 0 (Phase lag) 10 + w

SOL 5.44

Correct option is (C). From the given bode plot we can analyze that: 1. Slope - 40 dB/decade"2 poles 2. Slope - 20 dB/decade (Slope changes by + 20 dB/decade)"1 Zero 3. Slope 0 dB/decade (Slope changes by + 20 dB/decade)"1 Zero So there are 2 poles and 2 zeroes in the transfer function.

SOL 5.45

Correct option is (C). Characteristic equation for the system K 1+ =0 s (s + 3) (s + 10) s (s + 3) (s + 10) + K = 0 s3 + 13s2 + 30s + K = 0 Applying Routh’s stability criteria s3

1

30


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Control Systems

K

s2

13

s1

(13 # 30) - K 13 K

s0

Page 325

For stability there should be no sign change in first column So,

390 - K > 0 & K < 390 K >0 0 < K < 90

SOL 5.46

Correct option is (C). Given transfer function is

nodia 100 s + 20s + 100 Characteristic equation of the system is given by H (s)) =

2

s2 + 20s + 100 = 0 wn2 = 100 & wn = 10 rad/sec. 2xwn = 20 or x = 20 = 1 2 # 10 (x = 1) so system is critically damped.

SOL 5.47

Correct option is (D). State space equation of the system is given by, o = AX + Bu X Y = CX Taking Laplace transform on both sides of the equations. sX (s) = AX (s) + BU (s) (sI - A) X (s) = BU (s) X (s) = (sI - A) - 1 BU (s) Since Y (s) = CX (s) so Y (s) = C (sI - A) - 1 BU (s) Y (s) = C (sI - A) - 1 B U (s) s 0 0 1 (sI - A) = > H - > 0 s 0 - 2H s -1 => 0 s + 2H 1 >s + 2 1H (sI - A) - 1 = s (s + 2) 0 s V R 1 W S1 s s (s + 2)W = SS 1 W S0 (s + 2) W Transfer function X T T.F =


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Control Systems

Chapter 5

V R 1 W S1 s s (s + 2)W 0 G (s) = C [sI - A] - 1 B = 81 0BSS 1 W>1H 0 S (s + 2) W TR 1 V X W S s (s + 2)W S = 81 0BS 1 W S (s + 2) W X 1T = s (s + 2) SOL 5.48

Correct option is (A). Steady state error is given by, sR (s) ess = lim = G s " 0 1 + G (s) H (s) R (s) = L [r (t)] = 1 (Unit step input) s 1 G (s) = s (s + 2)

nodia Here

So,

SOL 5.49

H (s) = 1 (Unity feed back) s ^ s1 h ess = lim > H 1 s"0 1 + s (s + 2) s (s + 2) = lim = G =0 s " 0 s (s + 2) + 1

Correct option is (D). For input u1 , the system is (u2 = 0)

System response is (s - 1) (s + 2) H1 (s) = (s - 1) 1 1+ (s + 2) (s - 1) (s - 1) (s + 3) Poles of the system is lying at s =- 3 (negative s -plane) so this is stable. For input u2 the system is (u1 = 0) =

System response is


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Control Systems

Page 327

1 (s - 1) H2 (s) = (s - 1) 1+ 1 (s - 1) (s + 2) (s + 2) = (s - 1) (s + 3) One pole of the system is lying in right half of s -plane, so the system is unstable. SOL 5.50

Correct option is (B). Given function is. 1 s (s + 1) (s + 2) 1 G (jw) = jw (1 + jw) (2 + jw) G (s) =

nodia By simplifying

- jw 1 - jw 2 - jw 1 1 G (jw) = c 1 # # # m c m c jw - jw 1 + jw 1 - jw 2 + jw 2 - jw m jw 1 - jw 2 - j w = c- 2 mc w 1 + w2 mc 4 + w2 m - jw (2 - w2 - j3w) = 2 w (1 + w2) (4 + w2) jw (w2 - 2) - 3w2 = 2 2 2 + 2 w (1 + w ) (4 + w ) w (1 + w2) (4 + w2) G (jw) = x + iy x = Re [G (jw)] w " 0 = - 3 =- 3 1#4 4 +

SOL 5.51

Correct option is (D). Let response of the un-compensated system is 900 H UC (s) = s (s + 1) (s + 9) Response of compensated system. 900 HC (s) = G (s) s (s + 1) (s + 9) C Where GC (s) " Response of compensator Given that gain-crossover frequency of compensated system is same as phase crossover frequency of un-compensated system So, (wg) compensated = (wp) uncompensated - 180c = +H UC (jwp) - 180c =- 90c - tan- 1 (wp) - tan- 1 a

wp 9k

J w + wp N p 9 O 90c = tan KK 2 O K 1 - wp O 9 P L w2p 1=0 9 -1

So,

wp = 3 rad/sec. (wg) compensated = 3 rad/sec.


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Control Systems

Chapter 5

At this frequency phase margin of compensated system is fPM = 180c + +HC (jwg) 45c = 180c - 90c - tan- 1 (wg) - tan- 1 (wg /9) + +GC (jwg) 45c = 180c - 90c - tan- 1 (3) - tan- 1 (1/3) + +GC (jwg) 3 + 13 + +GC (jw g) 45c = 90c - tan-1 > 1 - 3 ^ 13 hH 45c = 90c - 90c + +GC (jwg) +GC (jwg) = 45c The gain cross over frequency of compensated system is lower than un-compensated system, so we may use lag-lead compensator. At gain cross over frequency gain of compensated system is unity so. HC (jwg) = 1

nodia 900 GC (jwg)

wg

wg2 + 1 wg2 + 81

=1

GC (jwg) = 3 9 + 1 9 + 81 900 = 3 # 30 = 1 900 10 in dB GC (wg) = 20 log b 1 l 10

=- 20 dB (attenuation)

SOL 5.52

Correct option is (B). Characteristic equation for the given system, K (s + 3) 1+ =0 (s + 8) 2 (s + 8) 2 + K (s + 3) = 0 s2 + (16 + K) s + (64 + 3K) = 0 By applying Routh’s criteria. s2 s1 s

0

64 + 3K

1 16 + K

0

64 + 3K

For system to be oscillatory 16 + K = 0 & K =- 16 Auxiliary equation A (s) = s2 + (64 + 3K) = 0 &

s2 + 64 + 3 # (- 16) = 0 s2 + 64 - 48 = 0 s2 =- 16 & jw = 4j w = 4 rad/sec

SOL 5.53

Correct option is (D). From the given block diagram we can obtain signal flow graph of the system. Transfer function from the signal flow graph is written as


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Control Systems

Page 329

c 0 P + c1 P s s2 T.F = 1 + a1 + a20 - Pb2 0 - Pb1 s s s s (c 0 + c1 s) P (s2 + a1 s + a 0) - P (b 0 + sb1) (c 0 + c1 s) P 2 ^s + a1 s + a 0h = P (b + sb1) 1- 2 0 s + a1 s + a 0 from the given reduced form transfer function is given by T.F = XYP 1 - YPZ by comparing above two we have =

nodia X = (c 0 + c1 s) 1 Y = 2 s + a1 s + a 0 Z = (b 0 + sb1)

SOL 5.54

Correct option is (A). For the given system Z is given by Z = E (s) Ki s

Where E (s) " steady state error of the system sR (s) Here E (s) = lim s " 0 1 + G (s) H (s) Input R (s) = 1 (Unit step) s

w2 G (s) = b Ki + K p le 2 s s + 2xws + w2 o H (s) = 1 (Unity feed back) So,

SOL 5.55

R V sb 1 l S W s S Wb Ki l Z = lim 2 s"0S K w W s i S1 + b s + K p l (s2 + 2xws + w2) W T X K i = Ki = 1 = lim 2 s"0 >s + (Ki + K p s) 2 w H Ki 2 (s + 2xws + w )

Correct option is (C). System response of the given circuit can be obtained as. 1 bCs l e 0 (s) H (s) = = 1 ei (s) bR + Ls + Cs l 1 H (s) = 2 LCs + RCs + 1 1 b LC l = s2 + R s + 1 L LC


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Control Systems

Characteristic equation is given by, s2 + R s + 1 = 0 L LC Here natural frequency wn = 1 LC R 2xwn = L Damping ratio x = R LC = R 2 2L x = 10 2 So peak overshoot is given by Here

% peak overshoot = e

SOL 5.57

C L

1 # 10- 3 = 0.5 (under damped) 10 # 10- 6

- px 1 - x2

# 100

nodia =e

SOL 5.56

Chapter 5

- p # 0.5 1 - (0.5) 2

# 100 = 16%


Correct option is (B). In standard form for a characteristic equation give as

sn + an - 1 sn - 1 + ... + a1 s + a 0 = 0 in its state variable representation matrix A is given as V R 1 0 g 0 W S 0 S 0 0 1 g 0 W A =S W Sh h h h h W S- a 0 - a1 - a2 g - an - 1W X T Characteristic equation of the system is

SOL 5.58

SOL 5.59

4s2 - 2s + 1 = 0 So, a2 = 4, a1 =- 2, a 0 = 1 R 0 1 0 VW RS 0 1 0 VW S A =S 0 0 1 W=S 0 0 1 W SS- a - a - a WW SS- 1 2 - 4WW 0 1 2 T X T X Correct option is (A). In the given options only in option (A) the nyquist plot does not enclose the unit circle (- 1, j0), So this is stable. Correct option is (A). Given function is, 10 4 (1 + jw) H (jw) = (10 + jw) (100 + jw) 2 Function can be rewritten as, 10 4 (1 + jw) H (jw) = 2 10 91 + j w C 10 4 91 + j w C 10 100 0.1 (1 + jw) = w w 2 a1 + j 10 ka1 + j 100 k


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Control Systems

Page 331

The system is type 0, So, initial slope of the bode plot is 0 dB/decade. Corner frequencies are w1 = 1 rad/sec w 2 = 10 rad/sec w 3 = 100 rad/sec As the initial slope of bode plot is 0 dB/decade and corner frequency w1 = 1 rad/ sec, the Slope after w = 1 rad/sec or log w = 0 is(0 + 20) =+ 20 dB/dec. After corner frequency w2 = 10 rad/sec or log w2 = 1, the Slope is (+ 20 - 20) = 0 dB/dec. Similarly after w3 = 100 rad/sec or log w = 2 , the slope of plot is (0 - 20 # 2) =- 40 dB/dec. SOL 5.60

nodia Correct option is (B). Given characteristic equation.

(s2 - 4) (s + 1) + K (s - 1) = 0 K (s - 1) or 1+ 2 =0 (s - 4) (s + 1) So, the open loop transfer function for the system. K (s - 1) , no. of poles n = 3 G (s) = (s - 2) (s + 2) (s + 1) no of zeroes m = 1 Steps for plotting the root-locus (1) Root loci starts at s = 2, s =- 1, s =- 2 (2) n > m , therefore, number of branches of root locus b = 3 (3) Angle of asymptotes is given by (2q + 1) 180c , q = 0, 1 n-m (2 # 0 + 1) 180c (1) = 90c (3 - 1) (2 # 1 + 1) 180c (2) = 270c (3 - 1) (4) The two asymptotes intersect on real axis at / Poles - / Zeroes x = n-m (- 1 - 2 + 2) - (1) = =- 1 3-1 (5) Between two open-loop poles s =- 1 and s =- 2 there exist a break away point. (s2 - 4) (s + 1) K =(s - 1) dK = 0 ds s =- 1.5

SOL 5.61

Correct option is (C). Closed loop transfer function of the given system is, s2 + 4 T (s) = (s + 1) (s + 4)


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Control Systems

T (jw) =

Chapter 5

(jw) 2 + 4 (jw + 1) (jw + 4)

If system output is zero T (jw) =

4 - w2 =0 ^ jw + 1h (jw + 4)

4 - w2 = 0 w2 = 4 & w = 2 rad/sec SOL 5.62

SOL 5.63

Correct option is (A). From the given plot we can see that centroid C (point of intersection) where asymptotes intersect on real axis) is 0. So for option (a) G (s) = K3 s / Poles - / Zeros = 0 - 0 = 0 Centroid = n-m 3-0

nodia Correct option is (A). Open loop transfer function is. (s + 1) G (s) = s2 jw + 1 G (jw) = - w2 Phase crossover frequency can be calculated as. +G (jwp) =- 180c tan- 1 (wp) =- 180c wp = 0 Gain margin of the system is. 1 G.M = G (jwp) =

SOL 5.64

=

1 w +1 w2p 2 p

w2p =0 w2p + 1

Correct option is (C). Characteristic equation for the given system 1 + G (s) H (s) = 0 (1 - s) 1+K =0 (1 + s) (1 + s) + K (1 - s) = 0 s (1 - K) + (1 + K) = 0 For the system to be stable, coefficient of characteristic equation should be of same sign. 1 - K > 0, K + 1 > 0 K < 1, K > - 1 -1 < K < 1


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Control Systems

Page 333

K <1 SOL 5.65

Correct option is (C). In the given block diagram

Steady state error is given as ess = lim sE (s)

nodia s"0

E (s) = R (s) - Y (s) Y (s) can be written as

Y (s) = :"R (s) - Y (s), 3 - R (s)D 2 s s+2

= R (s) ; 6 - 2 - Y (s) ; 6 E s (s + 2) s + 2 E s (s + 2) 6 Y (s) ;1 + = R (s) ; 6 - 2s E E s (s + 2) s (s + 2) (6 - 2s) Y (s) = R (s) 2 (s + 2s + 6) (6 - 2s) So, R (s) E (s) = R (s) - 2 (s + 2s + 6) 2 = R (s) ; 2 s + 4s E s + 2s + 6 For unit step input R (s) = 1 s Steady state error ess = lim sE (s) s"0

(s2 + 4s) ess = lim =s 1 2 =0 s (s + 2s + 6)G s"0 SOL 5.66

Correct option is (B). When it passes through negative real axis at that point phase angle is - 180cSo +G (jw) H (jw) =- 180c - 0.25jw - p =- p 2 - 0.25jw =- p 2 j0.25w = p 2 p jw= 2 # 0.25 s = jw = 2p Put s = 2p in given open loop transfer function we get - 0.25 # 2p =- 0.5 G (s) H (s) s = 2p = pe 2p


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Control Systems

Chapter 5

So it passes through (- 0.5, j0) SOL 5.67

Correct option is (C). Open loop transfer function of the system is given by. G (s) H (s) = (K + 0.366s) ; 1 E s (s + 1) K + j0.366w G (jw) H (jw) = jw (jw + 1) Phase margin of the system is given as fPM = 60c = 180c + +G (jwg) H (jwg) Where wg " gain cross over frequency = 1 rad/sec 0.366wg So, - 90c - tan- 1 (wg) 60c = 180c + tan- 1 b K l = 90c + tan- 1 b 0.366 l - tan- 1 (1) K = 90c - 45c + tan- 1 b 0.366 l K 15c = tan- 1 b 0.366 l K 0.366 = tan 15c K K = 0.366 0.267

nodia = 1.366

SOL 5.68

Correct option is (A). Given state equation. o (t) = >0 1 H X (t) + >1H u (t) X 0 -3 0 0 1 1 Here A => ,B = > H H 0 -3 0 State transition matrix is given by, f (t) = L- 1 [(sI - A) - 1] s 0 0 1 s -1 [sI - A] = > H - > => H 0 s 0 -3 0 s + 3H 1 >s + 3 1H [sI - A] - 1 = s (s + 3) 0 s V R 1 W S1 s s (s + 3)W = SS 1 W S0 (s + 3) W X T f (t) = L- 1 [(sI - A) - 1] 1 => 0

SOL 5.69

1 3

(1 - e-3t) H e-3t

Correct option is (C). State transition equation is given by


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Control Systems

Page 335

X (s) = F (s) X (0) + F (s) BU (s) Here F (s) " state transition matrix V R 1 W S1 s s (s + 3)W F (s) = SS 1 W S0 (s + 3) W X X (0) " initial conditionT -1 X (0) = > H 3 1 B => H 0 V R R1 1 W 1 VW S1 S s s (s + 3)W - 1 Ss (s + 3) s W 1 1 So X (s) = SS 1 W> 3 H + S 1 W>0H s 0 0 S (s + 3) W S s+3 W X T X RT 1 3 VW S- + s s (s + 3)W 1 1 + s =S S 0+ 3 W >0H s S s+3 W V TR X S- 1 W 1 s+3 = S 3 W + >s2 H W S 0 S s+3 W RT X V S 12 - 1 W s + 3W X (s) = Ss S W 3 S s+3 W Taking inverse Laplace Ttransform,Xwe get state transition equation as,

nodia t - e- 3t X (t) = > - 3t H 3e SOL 5.70

Option () is correct Phase margin of a system is the amount of additional phase lag required to bring the system to the point of instability or (- 1, j0) So here phase margin = 0c

SOL 5.71

Correct option is (D). Given transfer function is 5 s (s + 3s + 2) 5 F (s) = s (s + 1) (s + 2) By partial fraction, we get 5 F (s) = 5 - 5 + 2s s + 1 2 (s + 2) Taking inverse Laplace of F (s) we have f (t) = 5 u (t) - 5e- t + 5 e- 2t 2 2 So, the initial value of f (t) is given by lim f (t) = 5 - 5 + 5 (1) = 0 2 2 t"0 F (s) =

2


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Control Systems

Chapter 5

Correct option is (C). In A.C techo-meter output voltage is directly proportional to differentiation of rotor displacement. e (t) \ d [q (t)] dt dq (t) e (t) = Kt dt Taking Laplace tranformation on both sides of above equation E (s) = Kt sq (s) So transfer function E (s) T.F = = ^Kt h s q (s)

SOL 5.73

nodia Correct option is (B). Given characteristic equation,

s3 - 4s2 + s + 6 = 0 Applying Routh’s method, s3 s2

1 -4

1

6

s1

- 4 - 6 = 2.5 -4

s0

6

0

There are two sign changes in the first column, so no. of right half poles is 2. No. of roots in left half of s -plane SOL 5.74

= (3 - 2) = 1

Correct option is (B). Block diagram of the system is given as.

From the figure we can see that C (s) = :R (s) 1 + R (s)D 1 + R (s) s s C (s) = R (s) : 12 + 1 + 1D s s 2 C (s) = 1 + s2+ s R (s) s SOL 5.75



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Control Systems

Page 337

Characteristic equation is given by, sI - A = 0 s 0 0 2 (sI - A) = > H - > H 0 s 2 0 s -2 => -2 s H = s2 - 4 = 0 s1, s2 = ! 2 SOL 5.76

Correct option is (D). For the given system, characteristic equation can be written as, 1 + K (1 + sP) = 0 s (s + 2)

nodia s (s + 2) + K (1 + sP) = 0 s2 + s (2 + KP) + K = 0

From the equation. wn = K = 5 rad/sec (given) So, K = 25 and 2xwn = 2 + KP 2 # 0.7 # 5 = 2 + 25P or P = 0.2 so K = 25 , P = 0.2

SOL 5.77

Correct option is (D). Unit - impulse response of the system is given as, c (t) = 12.5e- 6t sin 8t , t $ 0 So transfer function of the system. H (s) = L [c (t)] = 12.52# 8 2 (s + 6) + (8) 100 H (s) = 2 s + 12s + 100 Steady state value of output for unit step input, lim y (t) = lim sY (s) = lim sH (s) R (s)

t"3

s"0

s"0

100 1 = lim s ; 2 s"0 s + 12s + 100 E s = 1.0 SOL 5.78

Correct option is (A). System response is. s s+1 jw H (jw) = jw + 1 H (s) =

Amplitude response H (jw) =

w w+1


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Control Systems

Chapter 5

Given input frequency w = 1 rad/sec. 1 So H (jw) w = 1 rad/sec = = 1 1+1 2 Phase response qh (w) = 90c - tan- 1 (w) qh (w) w = 1 = 90c - tan- 1 (1) = 45c So the output of the system is y (t) = H (jw) x (t - qh) = 1 sin (t - 45c) 2 SOL 5.79

Correct option is (C). Given open loop transfer function jaw + 1 G (jw) = (jw) 2 Gain crossover frequency (wg) for the system.

nodia G (jwg) = 1

a2 wg2 + 1 =1 - wg2

a2 wg2 + 1 = wg4 wg4 - a2 wg2 - 1 = 0 Phase margin of the system is

...(1)

fPM = 45c = 180c + +G (jwg) 45c = 180c + tan- 1 (wg a) - 180c tan- 1 (wg a) = 45c wg a = 1 From equation (1) and (2) 1 -1-1 = 0 a4 a 4 = 1 & a = 0.841 2 SOL 5.80

...(2)

Correct option is (C). Given system equation is. d 2 x + 6 dx + 5x = 12 (1 - e- 2t) dt dt 2 Taking Laplace transform on both side. s2 X (s) + 6sX (s) + 5X (s) = 12 :1 - 1 D s s+2 (s2 + 6s + 5) X (s) = 12 ; 2 E s (s + 2) System transfer function is 24 X (s) = s (s + 2) (s + 5) (s + 1) Response of the system as t " 3 is given by lim f (t) = lim sF (s) (final value theorem)

t"3

s"0


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Control Systems

Page 339

24 = lim s ; s"0 s (s + 2) (s + 5) (s + 1)E = 24 = 2.4 2#5 SOL 5.81

Correct option is (A). Transfer function of lead compensator is given by. K a1 + s k a H (s) = s a1 + b k R w V S1 + j a a kW H (jw) = K S W SS1 + j a w kWW b T So, phase response of the compensator is. X

nodia qh (w) = tan- 1 a w k - tan- 1 a w k a b Jw - wN -1 K a b O = tan- 1 w (b - a) = tan ; ab + w2 E 2 O KK 1+ w O ab P L qh should be positive for phase lead compensation w (b - a) So, >0 qh (w) = tan- 1 ; ab + w2 E b >a

SOL 5.82

Correct option is (A). Since there is no external input, so state is given by X (t) = f (t) X (0) f (t) "state transition matrix X [0] "initial condition e- 2t 0 2 So x (t) = > H> H 0 e- t 3 2e- 2t x (t) = > - t H 3e At t = 1, state of the system 2e- 2 x (t) t = 1 = > - 1H 2e 0.271 => 1.100H

SOL 5.83

Correct option is (B). Given equation d2 x + 1 dx + 1 x = 10 + 5e- 4t + 2e- 5t dt2 2 dt 18 Taking Laplace on both sides we have s2 X (s) + 1 sX (s) + 1 X (s) = 10 + 5 + 2 2 18 s s+4 s+5 10 (s + 4) (s + 5) + 5s (s + 5) + 2s (s + 4) (s2 + 1 s + 1 ) X (s) = 2 18 s (s + 4) (s + 5)


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Control Systems

Chapter 5

System response is, 10 (s + 4) (s + 5) + 5s (s + 5) + 2s (s + 4) s (s + 4) (s + 5) bs2 + 1 s + 1 l 2 18 10 (s + 4) (s + 5) + 5s (s + 5) + 2s (s + 4) = s (s + 4) (s + 5) bs + 1 lbs + 1 l 3 6 We know that for a system having many poles, nearness of the poles towards imaginary axis in s -plane dominates the nature of time response. So here time constant given by two poles which are nearest to imaginary axis. Poles nearest to imaginary axis s1 =- 1 , s2 =- 1 3 6 X (s) =

SOL 5.84

t1 = 3 sec

nodia So, time constants )

t2 = 6 sec

Correct option is (A). Steady state error for a system is given by sR (s) ess = lim s " 0 1 + G (s) H (s) 1 Where input R (s) = (unit step) s G (s) = b 3 lb 15 l s + 15 s + 1 H (s) = 1

So

(unity feedback) s ^ s1 h ess = lim 45 s"0 1+ (s + 15) (s + 1) = 15 = 15 60 15 + 45 %ess = 15 # 100 60 = 25%

SOL 5.85

Correct option is (C). Characteristic equation is given by 1 + G (s) H (s) = 0 Here

So,

H (s) = 1

(unity feedback) G (s) = b 3 lb 15 l s + 15 s + 1 1 + b 3 lb 15 l = 0 s + 15 s + 1 (s + 15) (s + 1) + 45 = 0 s2 + 16s + 60 = 0 (s + 6) (s + 10) = 0 s =- 6, - 10

SOL 5.86

Correct option is (A). Given equation can be written as,


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Control Systems

Page 341

d 2 w =- b dw - K 2 w + K V LJ a J dt LJ dt 2 Here state variables are defined as, dw = x 1 dt w = x2 So state equation is 2 xo1 =- B x1 - K x2 + K Va J LJ LJ xo2 = dw = x1 dt

In matrix form

K/LJ xo1 - B/J - K 2 /LJ x1 >o H = > >x H + > 0 H Va H x2 1 0 2 R 2 V Sd w W S dt2 W = P >dwH + QVa dt S dw W S dt W T X

nodia So matrix P is

- B/J - K 2 /LJ > 1 H 0

SOL 5.87

Correct option is (C). Characteristic equation of the system is given by 1 + GH = 0 K =0 1+ s (s + 2) (s + 4) s (s + 2) (s + 4) + K = 0 s3 + 6s2 + 8s + K = 0 Applying routh’s criteria for stability s3

1

8 K

s1

6 K - 48 6

s0

K

s

2

System becomes unstable if K - 48 = 0 & K = 48 6 SOL 5.88

Correct option is (A). The maximum error between the exact and asymptotic plot occurs at corner frequency. Here exact gain(dB) at w = 0.5a is given by 2

1 + w2 a (0.5a) 2 1/2 = 20 log K - 20 log ;1 + E a2 = 20 log K - 0.96

gain(dB) w = 0.5a = 20 log K - 20 log


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Control Systems

Chapter 5

Gain(dB) calculated from asymptotic plot at w = 0.5a is = 20 log K = 20 log K - (20 log K - 0.96) dB = 0.96 dB Similarly exact phase angle at w = 0.5a is. qh (w) w = 0.5a =- tan- 1 a w k a =- tan- 1 b 0.5a l a

Error in gain (dB)

=- 26.56c Phase angle calculated from asymptotic plot at (w = 0.5a) is - 22.5c Error in phase angle =- 22.5 - (- 26.56c) = 4.9c

SOL 5.89

nodia Correct option is (B). Given block diagram

Given block diagram can be reduced as

^ s1 h = 1 s+3 1 + ^ s1 h3 1 bs l G2 = = 1 1 s + 12 1 + b l 12 s Further reducing the block diagram. Where

G1 =

2G 1 G 2 1 + (2G1 G2) 9 (2) b 1 lb 1 l s + 3 s + 12 = 1 + (2) b 1 lb 1 l (9) s + 3 s + 12

Y (s) =


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Control Systems

Page 343

2 (s + 3) (s + 12) + 18 2 = 2 s + 15s + 54 2 1 = = s (s + 9) (s + 6) 27 a1 + ka1 + s k 9 6 =

***********

nodia


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Electronics and Electrical Mesurements

Chapter 6

EE 6 ELECTRONICS AND ELECTRICAL MESUREMENTS


EE SP 6.2

ONE MARK

Power consumed by a balanced 3-phase, 3-wire load is measured by the two wattmeter method. The first wattmeter reads twice that of the second. Then, the load impedance angle in radians is (A) p/12 (B) p/8 (C) p/6 (D) p/3

nodia In an oscilloscope screen, linear sweep is applied at the (A) vertical axis (B) horizontal axis (C) origin

(D) both horizontal and vertical axis YEAR 2014 EE01

TWO MARK

EE SP 6.3

An incandescent lamp is marked 40 W, 240 V. If resistance at room temperature ^26cCh is 120 W , and temperature coefficient of resistance is 4.8 # 10-3 /cC , then its ‘ON’ state filament temperature in cC is approximately_____.

EE SP 6.4

The reading of the voltmeter (rms) in volts, for the circuit shown in the figure is _________

EE SP 6.5

The dc current flowing in a circuit is measured by two ammeters, one PMMC and another electrodynamometer type, connected in series. The PMMC meter contains 100 turns in the coil, the flux density in the air gap is 0.2 Wb/m2 , and the area of the coil is 80 mm2 . The electrodynamometer ammeter has a change in mutual inductance with respect to deflection of 0.5 mH/deg. The spring constants of both the meters are equal. The value of current, at which the


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Page 345

deflections of the two meters are same, is ______ YEAR 2014 EE02 EE SP 6.6

EE SP 6.7

ONE MARK

The saw-tooth voltage waveform shown in the figure is fed to a moving iron voltmeter. Its reading would be close to __________

nodia

While measuring power of a three-phase balanced load by the two-wattmeter method, the readings are 100 W and 250 W. The power factor of the load is ______. YEAR 2014 EE02

EE SP 6.8

TWO MARKS

The total power dissipated in the circuit, shown in the figure, is 1 kW.

The voltmeter, across the load, reads 200 V. The value of XL is _____. EE SP 6.9

Suppose that resistors R1 and R2 are connected in parallel to give an equivalent resistor R. If resistors R1 and R2 have tolerance of 1% each, the equivalent resistor R for resistors R1 = 300 W and R2 = 200 W will have tolerance of (A) 0.5% (B) 1% (C) 1.2%

EE SP 6.10

(D) 2%

Two ammeters X and Y have resistances of 1.2 W and 1.5 W respectively and they give full-scale deflection with 150 mA and 250 mA respectively. The ranges have been extended by connecting shunts so as to give full scale deflection with 15 A. The ammeters along with shunts are connected in parallel and then placed in a circuit in which the total current flowing is 15 A. The current in amperes indicated in ammeter X is_____. YEAR 2014 EE03

EE SP 6.11

ONE MARK

An LPF wattmeter of power factor 0.2 is having three voltage settings 300 V, 150


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Chapter 6

V and 75 V, and two current settings 5 A and 10 A. The full scale reading is 150. If the wattmeter is used with 150 V voltage setting and 10 A current setting, the multiplying factor of the wattmeter is _______. EE SP 6.12

The two signals S1 and S2, shown in figure, are applied to Y and X deflection plates of an oscilloscope.

nodia The waveform displayed on the screen is


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YEAR 2014 EE03

Page 347

TWO MARKS

EE SP 6.13

A periodic waveform observed across a load is represented by 1 + sin wt 0 # wt < 6p V^t h = ) - 1 + sin wt 6p # wt < 12p The measured value, using moving iron voltmeter connected across the load, is 2 3 (B) (A) 3 2 (C) 3 (D) 2 2 3

EE SP 6.14

In the bridge circuit shown, the capacitors are loss free. At balance, the value of capacitance C1 in microfarad is ______.


ONE MARK

Three moving iron type voltmeters are connected as shown below. Voltmeter readings are V , V1 and V2 as indicated. The correct relation among the voltmeter readings is

(A) V = V1 + V2 2 2 (C) V = V1 V2 EE SP 6.16

(B) V = V1 + V2 (D) V = V2 - V1

The input impedance of the permanent magnet moving coil (PMMC) voltmeter is infinite. Assuming that the diode shown in the figure below is ideal, the reading of the voltmeter in Volts is


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(A) 4.46

(B) 3.15

(C) 2.23

(D) 0

nodia YEAR 2013

EE SP 6.17

EE SP 6.18

TWO MARKS

Two magnetically uncoupled inductive coils have Q factors q1 and q 2 at the chosen operating frequency. Their respective resistances are R1 and R2 . When connected in series, their effective Q factor at the same operating frequency is (A) q1 + q2 (B) ^1/q1h + ^1/q2h (C) ^q1 R1 + q2 R2h / ^R1 + R2h (D) ^q1 R2 + q2 R1h / ^R1 + R2h A strain gauge forms one arm of the bridge shown in the figure below and has a nominal resistance without any load as Rs = 300 W . Other bridge resistances are R1 = R2 = R 3 = 300 W . The maximum permissible current through the strain gauge is 20 mA. During certain measurement when the bridge is excited by maximum permissible voltage and the strain gauge resistance is increased by 1% over the nominal value, the output voltage V0 in mV is

(A) 56.02 (C) 29.85

(B) 40.83 (D) 10.02


Chapter 6

ONE MARK

A periodic voltage waveform observed on an oscilloscope across a load is shown. A permanent magnet moving coil (PMMC) meter connected across the same load reads


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(A) 4 V (C) 8 V EE SP 6.20

(B) 5 V (D) 10 V

nodia The bridge method commonly used for finding mutual inductance is (A) Heaviside Campbell bridge (B) Schering bridge (C) De Sauty bridge

EE SP 6.21

(D) Wien bridge

For the circuit shown in the figure, the voltage and current expressions are v (t) = E1 sin (wt) + E 3 sin (3wt) and i (t) = I1 sin (wt - f1) + I 3 sin (3wt - f3) + I5 sin (5wt) The average power measured by the wattmeter is

(A) 1 E1 I1 cos f1 2 (C) 1 [E1 I1 cos f1 + E 3 I 3 cos f3] 2

(B) 1 [E1 I1 cos f1 + E1 I 3 cos f3 + E1 I5] 2 (D) 1 [E1 I1 cos f1 + E 3 I1 cos f1] 2


TWO MARKS

An analog voltmeter uses external multiplier settings. With a multiplier setting of 20 kW, it reads 440 V and with a multiplier setting of 80 kW, it reads 352 V. For a multiplier setting of 40 kW, the voltmeter reads (A) 371 V (B) 383 V (C) 394 V (D) 406 V YEAR 2011

EE SP 6.23

Page 349

ONE MARK

A dual trace oscilloscope is set to operate in the ALTernate mode. The control input of the multiplexer used in the y -circuit is fed with a signal having a


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Chapter 6

frequency equal to (A) the highest frequency that the multiplexer can operate properly (B) twice the frequency of the time base (sweep) oscillator (C) the frequency of the time base (sweep) oscillator (D) haif the frequency of the time base (sweep) oscillator EE SP 6.24

Consider the following statement (1) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil. (2) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the voltage coil circuit. (A) (1) is true but (2) is false (C) both (1) and (2) are true

EE SP 6.25

(B) (1) is false but (2) is true (D) both (1) and (2) are false

nodia

The bridge circuit shown in the figure below is used for the measurement of an unknown element ZX . The bridge circuit is best suited when ZX is a

(A) low resistance (C) low Q inductor

(B) high resistance (D) lossy capacitor


TWO MARKS

A 4 12 digit DMM has the error specification as: 0.2% of reading + 10 counts. If a dc voltage of 100 V is read on its 200 V full scale, the maximum error that can be expected in the reading is (A) ! 0.1% (B) ! 0.2% (C) ! 0.3% (D) ! 0.4% YEAR 2010

EE SP 6.27

ONE MARK

A wattmeter is connected as shown in figure. The wattmeter reads.


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Page 351

(A) Zero always (B) Total power consumed by Z1 and Z 2 (C) Power consumed by Z1 (D) Power consumed by Z2 EE SP 6.28

nodia

An ammeter has a current range of 0-5 A, and its internal resistance is 0.2 W. In order to change the range to 0-25 A, we need to add a resistance of (A) 0.8 W in series with the meter (B) 1.0 W in series with the meter (C) 0.04 W in parallel with the meter (D) 0.05 W in parallel with the meter

EE SP 6.29

As shown in the figure, a negative feedback system has an amplifier of gain 100 with ! 10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The overall system gain is approximately :

(B) 10 ! 2% (D) 10 ! 10%

(A) 10 ! 1% (C) 10 ! 5% YEAR 2010 EE SP 6.30

TWO MARKS

The Maxwell’s bridge shown in the figure is at balance. The parameters of the inductive coil are.

(A) R = R2 R 3 /R 4, L = C 4 R2 R 3


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Chapter 6

(B) L = R2 R 3 /R 4, R = C 4 R2 R 3 (C) R = R 4 /R2 R 3, L = 1/ (C 4 R2 R 3) (D) L = R 4 /R2 R 3, R = 1/ (C 4 R2 R 3) YEAR 2009 EE SP 6.31

ONE MARK

The pressure coil of a dynamometer type wattmeter is (A) Highly inductive (B) Highly resistive (C) Purely resistive

EE SP 6.32

(D) Purely inductive

The two inputs of a CRO are fed with two stationary periodic signals. In the X-Y mode, the screen shows a figure which changes from ellipse to circle and back to ellipse with its major axis changing orientation slowly and repeatedly. The following inference can be made from this. (A) The signals are not sinusoidal

nodia (B) The amplitudes of the signals are very close but not equal

(C) The signals are sinusoidal with their frequencies very close but not equal (D) There is a constant but small phase difference between the signals YEAR 2009

EE SP 6.33

The figure shows a three-phase delta connected load supplied from a 400V, 50 Hz, 3-phase balanced source. The pressure coil (PC) and current coil (CC) of a wattmeter are connected to the load as shown, with the coil polarities suitably selected to ensure a positive deflection. The wattmeter reading will be

(A) 0 (C) 800 Watt EE SP 6.34

TWO MARKS

(B) 1600 Watt (D) 400 Watt

An average-reading digital multi-meter reads 10 V when fed with a triangular wave, symmetric about the time-axis. For the same input an rms-reading meter will read (B) 10 (A) 20 3 3 (C) 20 3 (D) 10 3 YEAR 2008

EE SP 6.35

ONE MARK

Two 8-bit ADCs, one of single slope integrating type and other of successive


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Page 353

approximate type, take TA and TB times to convert 5 V analog input signal to equivalent digital output. If the input analog signal is reduced to 2.5 V, the approximate time taken by the two ADCs will respectively, be (B) TA /2, TB (A) TA, TB (C) TA, TB /2 (D) TA /2, TB /2 YEAR 2008 EE SP 6.36

TWO MARKS

Two sinusoidal signals p (w1, t) = A sin w1 t and q (w2 t) are applied to X and Y inputs of a dual channel CRO. The Lissajous figure displayed on the screen shown below :

nodia The signal q (w2 t) will be represented as (A) q (w2 t) = A sin w2 t, w2 = 2w1 (B) q (w2 t) = A sin w2 t, w2 = w1 /2 (C) q (w2 t) = A cos w2 t, w2 = 2w1 (D) q (w2 t) = A cos w2 t, w2 = w1 /2 EE SP 6.37

The ac bridge shown in the figure is used to measure the impedance Z .

If the bridge is balanced for oscillator frequency f = 2 kHz, then the impedance Z will be (A) (260 + j0) W (B) (0 + j200) W (C) (260 - j200) W (D) (260 + j200) W YEAR 2007 EE SP 6.38

ONE MARK

The probes of a non-isolated, two channel oscillocope are clipped to points A,


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Chapter 6

B and C in the circuit of the adjacent figure. Vin is a square wave of a suitable low frequency. The display on Ch1 and Ch2 are as shown on the right. Then the “Signal” and “Ground” probes S1, G1 and S2, G2 of Ch1 and Ch2 respectively are connected to points :

nodia (A) A, B, C, A

(B) A, B, C, B

(C) C, B, A, B

(D) B, A, B, C

YEAR 2007

EE SP 6.39

TWO MARKS

A bridge circuit is shown in the figure below. Which one of the sequence given below is most suitable for balancing the bridge ?

(A) First adjust (B) First adjust (C) First adjust (D) First adjust

R4 , and then adjust R1 R2 , and then adjust R3 R2 , and then adjust R4 R4 , and then adjust R2


ONE MARK

The time/div and voltage/div axes of an oscilloscope have been erased. A student connects a 1 kHz, 5 V p-p square wave calibration pulse to channel-1 of the scope and observes the screen to be as shown in the upper trace of the figure. An unknown signal is connected to channel-2(lower trace) of the scope. It the time/ div and V/div on both channels are the same, the amplitude (p-p) and period of the unknown signal are respectively


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EE SP 6.41


(A) 5 V, 1 ms

(B) 5 V, 2 ms

(C) 7.5 V, 2 ms

(D) 10 V, 1 ms

Page 355

A sampling wattmeter (that computes power from simultaneously sampled values of voltage and current) is used to measure the average power of a load. The peak to peak voltage of the square wave is 10 V and the current is a triangular wave of 5 A p-p as shown in the figure. The period is 20 ms. The reading in W will be

nodia (A) 0 W (C) 50 W

(B) 25 W (D) 100 W

YEAR 2006

TWO MARKS

EE SP 6.42

A current of - 8 + 6 2 (sin wt + 30%) A is passed through three meters. They are a centre zero PMMC meter, a true rms meter and a moving iron instrument. The respective reading (in A) will be (A) 8, 6, 10 (B) 8, 6, 8 (C) - 8 ,10,10 (D) - 8 ,2,2

EE SP 6.43

A variable w is related to three other variables x ,y ,z as w = xy/z . The variables are measured with meters of accuracy !0.5% reading, !1% of full scale value and !1.5% reading. The actual readings of the three meters are 80, 20 and 50 with 100 being the full scale value for all three. The maximum uncertainty in the measurement of w will be (A) !0.5% rdg (B) !5.5% rdg (C) !6.7 rdg (D) !7.0 rdg

EE SP 6.44

A 200/1 Current transformer (CT) is wound with 200 turns on the secondary on a toroidal core. When it carries a current of 160 A on the primary, the ratio and phase errors of the CT are found to be - 0.5% and 30 minutes respectively. If the number of secondary turns is reduced by 1 new ratio-error(%) and phase-


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error(min) will be respectively (A) 0.0, 30 (C) - 1.0, 30 EE SP 6.45

Chapter 6

(B) - 0.5, 35 (D) - 1.0, 25

R1 and R4 are the opposite arms of a Wheatstone bridge as are R3 and R2 . The source voltage is applied across R1 and R3 . Under balanced conditions which one of the following is true (B) R1 = R2 R3 /R4 (A) R1 = R3 R4 /R2 (D) R1 = R2 + R3 + R4 (C) R1 = R2 R4 /R3 YEAR 2005

EE SP 6.46

EE SP 6.47

nodia The Q-meter works on the principle of (A) mutual inductance

(B) self inductance

(C) series resonance

(D) parallel resonance

A PMMC voltmeter is connected across a series combination of DC voltage source V1 = 2 V and AC voltage source V2 (t) = 3 sin (4t) V. The meter reads (A) 2 V (B) 5 V (C) (2 +

EE SP 6.48

ONE MARK

3 /2) V

(D) ( 17 /2) V

A digital-to-analog converter with a full-scale output voltage of 3.5 V has a resolution close to 14 mV. Its bit size is (A) 4 (B) 8 (C) 16 (D) 32 YEAR 2005

TWO MARKS

EE SP 6.49

The simultaneous application of signals x (t) and y (t) to the horizontal and vertical plates, respectively, of an oscilloscope, produces a vertical figure-of-8 display. If P and Q are constants and x (t) = P sin (4t + 30c), then y (t) is equal to (A) Q sin (4t - 30c) (B) Q sin (2t + 15c) (C) Q sin (8t + 60c) (D) Q sin (4t + 30c)

EE SP 6.50

A DC ammeter has a resistance of 0.1 W and its current range is 0-100 A. If the range is to be extended to 0-500 A, then meter required the following shunt resistance (A) 0.010 W (B) 0.011 W (C) 0.025 W (D) 1.0 W

EE SP 6.51

The set-up in the figure is used to measure resistance R .The ammeter and voltmeter resistances are 0.01W and 2000 W, respectively. Their readings are 2 A and 180 V, respectively, giving a measured resistances of 90 W The percentage error in the measurement is


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EE SP 6.52


(A) 2.25%

(B) 2.35%

(C) 4.5%

(D) 4.71%

Page 357

A 1000 V DC supply has two 1-core cables as its positive and negative leads : their insulation resistances to earth are 4 MW and 6 MW, respectively, as shown in the figure. A voltmeter with resistance 50 kW is used to measure the insulation of the cable. When connected between the positive core and earth, then voltmeter reads

nodia (A) 8 V (C) 24 V

EE SP 6.53

(B) 16 V (D) 40 V

Two wattmeters, which are connected to measure the total power on a three-phase system supplying a balanced load, read 10.5 kW and - 2.5 kW, respectively. The total power and the power factor, respectively, are (A) 13.0 kW, 0.334 (B) 13.0 kW, 0.684 (C) 8.0 kW, 0.52 (D) 8.0 kW, 0.334 YEAR 2004

ONE MARK

EE SP 6.54

A dc potentiometer is designed to measure up to about 2 V with a slide wire of 800 mm. A standard cell of emf 1.18 V obtains balance at 600 mm. A test cell is seen to obtain balance at 680 mm. The emf of the test cell is (A) 1.00 V (B) 1.34 V (C) 1.50 V (D) 1.70 V

EE SP 6.55

The circuit in figure is used to measure the power consumed by the load. The current coil and the voltage coil of the wattmeter have 0.02 W and 1000W resistances respectively. The measured power compared to the load power will be


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EE SP 6.56


(A) 0.4 % less

(B) 0.2% less

(C) 0.2% more

(D) 0.4% more

A galvanometer with a full scale current of 10 mA has a resistance of 1000 W. The multiplying power (the ratio of measured current to galvanometer current) of 100 W shunt with this galvanometer is (A) 110 (B) 100

nodia (C) 11

(D) 10

YEAR 2004

EE SP 6.57

Chapter 6

TWO MARKS

A CRO probe has an impedance of 500 kW in parallel with a capacitance of 10 pF. The probe is used to measure the voltage between P and Q as shown in figure. The measured voltage will be

(A) 3.53 V (C) 4.54 V

(B) 4.37 V (D) 5.00 V

EE SP 6.58

A moving coil of a meter has 100 turns, and a length and depth of 10 mm and 20 mm respectively. It is positioned in a uniform radial flux density of 200 mT. The coil carries a current of 50 mA. The torque on the coil is (A) 200 mNm (B) 100 mNm (C) 2 mNm (D) 1 mNm

EE SP 6.59

A dc A-h meter is rated for 15 A, 250 V. The meter constant is 14.4 A-sec/rev. The meter constant at rated voltage may be expressed as (A) 3750 rev/kWh (B) 3600 rev/kWh (C) 1000 rev/kWh (D) 960 rev/kWh

EE SP 6.60

A moving iron ammeter produces a full scale torque of 240 mNm with a deflection of 120c at a current of 10 A . The rate of change of self induction (mH/radian) of the instrument at full scale is (A) 2.0 mH/radian (B) 4.8 mH/radian


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Page 359

(C) 12.0 mH/radian (D) 114.6 mH/radian EE SP 6.61

A single-phase load is connected between R and Y terminals of a 415 V, symmetrical, 3-phase, 4-wire system with phase sequence RYB. A wattmeter is connected in the system as shown in figure. The power factor of the load is 0.8 lagging. The wattmeter will read

nodia (A) - 795 W (C) + 597 W

EE SP 6.62

EE SP 6.63

(B) - 597 W (D) + 795 W

A 50 Hz, bar primary CT has a secondary with 500 turns. The secondary supplies 5 A current into a purely resistive burden of 1 W. The magnetizing ampere-turns is 200. The phase angle between the primary and second current is (A) 4.6c (B) 85.4c (C) 94.6c (D) 175.4c The core flux in the CT of Prob Q.44, under the given operating conditions is (A) 0 (B) 45.0 mWb (C) 22.5 mWb (D) 100.0 mWb YEAR 2003

ONE MARK

EE SP 6.64

A Manganin swap resistance is connected in series with a moving coil ammeter consisting of a milli-ammeter and a suitable shunt in order to (A) minimise the effect of temperature variation (B) obtain large deflecting torque (C) reduce the size of the meter (D) minimise the effect of stray magnetic fields

EE SP 6.65

The effect of stray magnetic field on the actuating torque of a portable instrument is maximum when the operating field of the instrument and the stray fields are (A) perpendicular (B) parallel (C) inclined at 60%

EE SP 6.66

(D) inclined at 30%

A reading of 120 is obtained when standard inductor was connected in the circuit of a Q-meter and the variable capacitor is adjusted to value of 300 pF. A lossless capacitor of unknown value Cx is then connected in parallel with the variable


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Chapter 6

capacitor and the same reading was obtained when the variable capacitor is readjusted to a value of 200 pF. The value of Cx in pF is (A) 100 (B) 200 (C) 300

(D) 500


EE SP 6.68

TWO MARKS

The simplified block diagram of a 10-bit A/D converter of dual slope integrator type is shown in figure. The 10-bit counter at the output is clocked by a 1 MHz clock. Assuming negligible timing overhead for the control logic, the maximum frequency of the analog signal that can be converted using this A/D converter is approximately

nodia (A) 2 kHz

(B) 1 kHz

(C) 500 Hz

(D) 250 Hz

The items in Group-I represent the various types of measurements to be made with a reasonable accuracy using a suitable bridge. The items in Group-II represent the various bridges available for this purpose. Select the correct choice of the item in Group-II for the corresponding item in Group-I from the following List-I P.

List-II

Resistance in the milli-ohm range

Q. Low values of Capacitance

1. Wheatstone Bridge 2. Kelvin Double Bridge

R. Comparison of resistance which are nearly equal 3. Schering Bridge S.

Inductance of a coil with a large time-constant 4. Wien’s Bridge 5. Hay’s Bridge 6. Carey-Foster Bridge

Codes : (A) P=2, Q=3, R=6, S=5 (B) P=2, Q=6, R=4, S=5 (C) P=2, Q= 3, R=5, S=4 (D) P=1, Q=3, R=2, S=6 EE SP 6.69

A rectifier type ac voltmeter of a series resistance Rs , an ideal full-wave rectifier bridge and a PMMC instrument as shown in figure. The internal. resistance of the instrument is 100 W and a full scale deflection is produced by a dc current of 1 mA. The value of Rs required to obtain full scale deflection with an ac voltage of 100 V (rms) applied to the input terminals is


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(A) 63.56 W (C) 89.93 W EE SP 6.70

EE SP 6.71

Page 361

(B) 69.93 W (D) 141.3 kW

A wattmeter reads 400 W when its current coil is connected in the R-phase and its pressure coil is connected between this phase and the neutral of a symmetrical 3-phase system supplying a balanced star connected 0.8 p.f. inductive load. This phase sequence is RYB. What will be the reading of this wattmeter if its pressure coil alone is reconnected between the B and Y phases, all other connections remaining as before ? (A) 400.0 (B) 519.6 (C) 300.0 (D) 692.8

nodia

The inductance of a certain moving-iron ammeter is expressed as L = 10 + 3q - (q2 /4) mH , where q is the deflection in radians from the zero position. The control spring torque is 25 # 10 - 6 Nm/radian. The deflection of the pointer in radian when the meter carries a current of 5 A, is (A) 2.4 (B) 2.0 (C) 1.2 (D) 1.0

EE SP 6.72

A 500A/5A, 50 Hz transformer has a bar primary. The secondary burden is a pure resistance of 1 W and it draws a current of 5 A. If the magnetic core requires 250 AT for magnetization, the percentage ratio error is (A) 10.56 (B) - 10.56 (C) 11.80 (D) - 11.80

EE SP 6.73

The voltage-flux adjustment of a certain 1-phase 220 V induction watt-hour meter is altered so that the phase angle between the applied voltage and the flux due to it is 85c(instead of 90c). The errors introduced in the reading of this meter when the current is 5 A at power factor of unity and 0.5 lagging are respectively (A) 3.8 mW, 77.4 mW (B) - 3.8 mW, - 77.4 mW (C) - 4.2 W, - 85.1 W (D) 4.2 W, 85.1 W

EE SP 6.74

Group-II represents the figures obtained on a CRO screen when the voltage signals Vx = Vxm sin wt and Vy = Vym sin (wt + F) are given to its X and Y plates respectively and F is changed. Choose the correct value of F from Group-I to match with the corresponding figure of Group-II.


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Group-I

Chapter 6

Group-II

P. F = 0

Q. F = p/2

R. p < F < 3p/2

nodia S. F = 3p/2

Codes : (A) P=1, Q= 3, R=6, S=5 (B) P=2, Q= 6, R=4, S=5 (C) P=2, Q= 3, R=5, S=4 (D) P=1, Q=5, R=6, S=4

***********


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Page 363

SOLUTIONS SOL 6.1

Correct option (C). For two wattmeter method, we define load impedance angle, f as tan f = 3 9w1 - w2 C w1 + w 2 where w1 and w2 are the readings of two wattmeters. Since, we have w1 = 2w2 So, we obtain the load impedance angle as tan f = 3 :2w2 - w2 D 2w2 + w2 = 3#1 = 1 3 3 or f = p rad 6

SOL 6.2

SOL 6.3

nodia

Correct option (B). During sweep time ^ts h, the beam moves from left to right across the CRT screen. The beam B deflected to the right by the increasing amplitude of the ramp voltage and the fact that the positive voltage attracts the negative electrons, and to make this effective we apply linear sweep in horizontal axis. Correct answer is 2470.44°C. For the incandescent lamp, we have P = 40 W , V = 240 V So, the resistance of its filament in ON state is 2 ^240h2 RQ = V = = 1440 W 40 P Now, we have R = 120 W

at t = 26cC ,

a = 4.5 # 10-3 /cC Since, the relation between resistance and temperature is given by and

RQ = R 61 + a ^Q2 - Q1h@ Substituting the given values, we get 1440 = 120 81 + 4.5 # 10-3 ^Q2 - 26hB 12 = 1 + 4.5 # 10-3 ^Q2 - 26h

or or or Thus, SOL 6.4

11 4.5 # 10-3 Q2 - 26 = 2.444.44 Q2 = 2470.44cC Q2 - 26 =

Correct answer is 141.42 V. We have the circuit,


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Chapter 6

It is a Wheatstone bridge. For the bridge to be balanced, the required condition is ...(1) z1 z 4 = z 2 z 3 From the given circuit, we have

nodia z1 = z 4 = j 1 W

z2 = z3 = 1 W j Substituting these values in equation (1), we get j#j = 1#1 j j j2 = 12 j - 1 =- 1 Hence, it is a balanced Wheatstone bridge, and reading of voltmeter (rms) is

and

V = 100 # 2 = 141.42 V SOL 6.5

Correct answer is 3.2 A. Given that the two ammeters, one PMMC and another electrodynamometer type, are connected in series. So, we have

For PMMC coil, we define the deflecting torque as t d = NBAI = GI

...(1)

where G is a constant, given as G = NBA Again, the controlling torque is defined as Tc = K q

...(2)

At balance, we have Tc = Td


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deflection q = G I [using equations (1) and (2)] K Now, for electrodynamometer, we have deflecting torque Td = I1 I2 dM dq So,

Page 365

...(3)

I1 = I2 = I (series)

where

Td = I 2 dM dq

So,

...(4)

The controlling torque is ...(5) Tc = Kq Therefore, equating the equations (4) and (5) for balance condtion, the steady deflection is obtained as 2 ...(6) q = I dM K dq Since, it is given in the question that deflection is same in both meter. So, from equations (3) and (6), we get G I = I 2 dM K dq K or (spring constants are equal) NBA = I dM dq

nodia or

Thus,

SOL 6.6

100 # 0.2 # 80 # 10-6 = I # 0.5 # 10-3 -3 I = 100 # 2 # 80 # 10 = 3.2 A 5

Correct answer is 57.74 . From the graph, we write mathematical expression of voltage v ^ t h 100 v^t h = t, 20 # 10-3 So, v ^ t h = 5 # 103 t Volt , A moving iron voltmeter reads rms value of voltage, so 1 T v2 t dt Vrms = ^ h T 0

#

= = = = = = SOL 6.7

1 20 # 10-3 25 # 106 20 # 10-3

#

20 # 10-3

#

20 # 10-3

0

0

2 ^5 # 103 # t h dt

t2 dt -3

25 # 10 t3 20 # 10 : D 20 # 10-3 3 0 3 ^20 # 10-3h 25 # 106 # 3 20 # 10-3 6 25 # 10 # 400 # 10-6 3 100 = 57.74 A 3 6

Correct answer is 0.8029. In two wattmeter method, power factor angle is given by f = tan-1 =

3 ^W1 - W2h G W1 + W2


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=

Chapter 6

3 ^250 - 100h 250 + 100

f = tan-1 ^0.7423h f = 36.586c Power factor = cos f

= cos ^36.586ch = 0.8029 lagging

SOL 6.8


nodia

Total power consumed, P = 1 KW = 1000 W We can see that power will be consumed by only 1 W resistor and load resistor R. From the figure we can notice that current in 1 W resistor is 2 A and in load resistorR is 10 A. So we can write total power equal to sum of power dissipated in these two resistors. P = (2) 2 (1) + (10) 2 R 1000 = 4 + 100R R = 9.96 W

Voltage across load,

VL = 200 V VL = I R2 + X L2

200 = 10 ^9.96h2 + X L2 ^20h2 = ^9.96h2 # X L2 XL = 400 - 99.20 XL = 17.34 W SOL 6.9

Correct option is (B). Equivalent resistance when connected in parallel is Req = R1 R2 R1 + R 2 Let % Tolerance in

R1 + R2 = Rsum ...(1) Req = TR1 ^%h + TR2 ^%h - TRsum ^%h TRsum = ^300 ! 1%h + ^200 ! 1%h = ;300 ! b 300 # 1 lE + ;200 ! b 200 # 1 lE 100 100 = ^300 ! 3h + ^200 ! 2h = 500 ! 5 TRsum ^%h = b 5 # 100 l 500 = 1%


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Now, from eq (1)

SOL 6.10

Page 367

TReq ^%h = 1% + 1% - 1% TReq ^%h = 1%

Correct answer is 10.16 . For ammeter X , Let shunt resistor be RmX

nodia ImX RmX = Î - ImX h RsnX 0.15 # 1.2 = ^15 - 0.15h RsnX RsnX = 0.18 = 0.0121 W 14.85 For ammeter Y , Let shunt resistor be RMY

IMY RMY = Î - IMY h RsnY 0.25 # 1.5 = ^15 - 0.25h RsnY RsnY = 0.375 = 0.0254 W 14.75 Equivalent resistance of ammeter X RX = RmX || RsnX = 1.2 || 0.0121 - 0.0121 W Similarly RY = 0.0254 W When connected in parallel, current in the circuit is 15 A , so current through ammeter X RY (Using current division) 15 IX = RX + RY # 0.0254 = 15 0.0121 + 0.0254 # = 10.16 A SOL 6.11

Correct answer is 2. In low power factor (LPF) wattmeter, the multiplying factor is given by power setting M.F. = power for maximum difflection


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Chapter 6

^voltage range usedh # ^current range usedh # (power factor) Full scale reading From the given data, we have Full scale reading = 150 Voltage range used = 150 V Current range used = 10 A Power factor = 0.2 Substituting these values in above expression, we get M.F. = 150 # 10 # 0.2 = 2 150 =

SOL 6.12

Correct option is (A). We consider the two signals S1 and S2 applied to Y and X deflection plates for the period T as

nodia

observing the waveforms, we condude that For 0 < t < T/2 , 0 < S1 < 1 and S2 = 1 At t = T , S1 = S2 = 0 2 For T < t < T , 0 > S1 > - 1 and S2 =- 1 2 Since, S1 is applied to Y -deflection plate and S2 is applied to X -deflection plate. So, we have For 0 < t < T/2 ; 0 < y < 1, x = 1 At t = T/2 ; x = y = 0 For T/2 < t < T ; - 1 < y < 0 , x =- 1 Thus, the waveform displayed on the CRT screen is


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SOL 6.13


Page 369

Correct option is (A). Moving iron instrument gives rms value, so measured value of moving iron is rms of given function 1 + sin wt 0 # wt # 6p v^t h = ) - 1 + sin wt 6p # wt # 12p So, we have

nodia 6p 1 + sin wt h2 d ^wt h + Vrms = = 1 ; 12p 0 ^

#

== 1 ; 12p 0

#

6p

^1 + 2 sin wt + sin2 wt h d ^wt h +

#

#

6p

12p

6p

12p

SOL 6.14

1/2

^1 - 2 sin wt + sin2 wt h d ^wt hEG

1 - cos 2wt d wt + = = 1 ;6p + 6p + ^ h 12p 2 0 1/2 = : 1 612p + 6p@D 12p 1/2 = b 18 l = 3 12 2

#

1/2

^- 1 + sin wt h2 d ^wt hEG

6p

#

12p

6p

1 - cos 2wt d wt 1/2 ^ hEG 2

Correct answer is 0.3 mF . Capacitor are lossless and bridge is balanced. On balancing, we have z1 z 4 = z 2 z 3 Now, the given bridge circuit is

For balancing, we have 35 # 103 # or SOL 6.15

1 1 3 -6 = 105 # 10 # w j #C jw # 0.1 # 10 C = 0.3 mF



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Chapter 6

For an ideal voltmeter interval resistance is always zero. So we can apply the KVL along the two voltmeters as V - V1 - V2 = 0 V = V1 + V2

or

SOL 6.16

nodia

Correct option is (A). For the + ve half cycle of I/p voltage, diode will be forward biased (Vg = 0 , ideal diode) Therefore, the voltmeter will be short circuited and reads (for + ve half cycle) V1 = 0 volt Now, for - ve half cycle, diode will be reverse biased and treated as open circuit. So, the voltmeter reads the voltage across 100 kW. Which is given by 14.14 0c V2 = 100 # 100 + 1 14 So, V2,rms = volt 2 Therefore, the average voltage for the whole time period is obtained as V + V2, rms 0 + ^14/ 2 h = = 14 Vave = 1 2 2 2 2 = 4.94 . 4.46 volt

SOL 6.17

Correct option is (C). The quality factor of the inductances are given by q1 = wL1 R1 and q2 = wL2 R2 So, in series circuit, the effective quality factor is given by XLeq = wL 1 + wL 2 Q = Req R1 + R 2 q1 q wL 1 + wL 2 + 2 q R + q2 R2 R R R R 2 2 1 R2 1 R2 = = 1 1 = 1 + 1 1 + 1 R1 + R 2 R 2 R1 R 2 R1

SOL 6.18


SOL 6.19

Correct option is (A). PMMC instrument reads average (dc) value. T 1 Vavg = 1 v (t) dt = T 0 20 # 10-3

#

10 = 1 ; tdt + 20 0

#

#

20

0

v (t) dt

20

12

# (- 5) dt + # 5dtE 10

12


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Page 371

2 10 20 = 1 c :t D - 5 6t @12 + 5 6t @12 m 10 20 2 0

= 1 [50 - 5 (2) + 5 (8)] = 80 = 4 V 20 20 SOL 6.20

Correct option is (A). Heaviside mutual inductance bridge measures mutual inductance is terms of a known self-inductance.

SOL 6.21

Correct option is (C). Let wt = q , we have instaneous voltage and current as follows. v (t) = E1 sin q + E 3 sin 3q i (t) = I1 sin (q - f1) + I 3 sin (3q - f3) + I5 sin (5q) We know that wattmeter reads the average power, which is gives as 2p P = 1 v (t) i (t) dq 2p 0 We can solve this integration using following results. 2p (1) 1 A sin (q + a):B sin (q + b) dq = 1 AB cos (a - b) 2p 0 2

nodia #

...(1)

#

2p

(2)

1 2p

#

(3)

1 2p

# A sin (mq + a):B cos (nq + b) dq = 0

(4)

1 2p

# A sin (mq + a):B cos (nq + b) dq = 0

0

A sin (q + a):B cos (q + a) dq = 1 AB sin (a - b) 2

2p

0

2p

0

Result (3) and (4) implies that power is transferred between same harmonics of voltages and currents. Thus integration of equation (1) gives. P = 1 E1 I1 cos f + 1 E 3 I 3 cos f3 2 2 SOL 6.22

Correct option is (D). A voltmeter with a multiplier is shown in figure below.

Here

Im = Fully scale deflection current of meter. Rm = Internal resistance of meter Rs = Voltage across the meter V = Full range voltage of instrument Vm = Im Rm V = Im (Rm + Rs)


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Chapter 6

V = R m + Rs = 1 + Rs Vm Rm Rm Rs1 = 20 kW , Vm1 = 440 V V = 1 + 20k 440 Rm

Here when, So,

Rs2 = 80 kW , Vm2 = 352 V V = 1 + 80 k So, 352 Rm Solving equation (1) and (2), we get

...(1)

When,

V = 480 V , Rm = 220 kW Rs3 = 40 kW , Vm3 = ? 480 = 1 + 40 k & V - 406 V m2 Vm3 220 k

So when

SOL 6.23

...(2)

nodia

Correct option is (C). In the alternate mode it switches between channel A and channel B, letting each through for one cycle of horizontal sweep as shown in the figure.

SOL 6.24

Correct option is (A). The compensating coil compensation the effect of impedance of current coil.

SOL 6.25

Correct option is (C). Let

Z1 = R1 || jwC1


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so admittance

Page 373

Y1 = 1 = 1 + jwC1 Z1 R 1

Z2 = R2 and Z 4 = R 4 Let ZX = RX + jwLX (Unknown impedance) For current balance condition of the Bridge Z 2 Z 4 = Z X Z1 = Z X Y1 Let

ZX = Z2 Z 4 Y1

RX + jwLX = R2 R 4 b 1 + jwC1 l R1 Equating imaginary and real parts RX = R2 R 4 and LX = R2 R 4 C1 R1 Quality factor of inductance which is being measured Q = wL X = wR 1 C 1 RX From above equation we can see that for measuring high values of Q we need a large value of resistance R 4 which is not suitable. This bridge is used for measuring low Q coils. Note: We can observe directly that this is a maxwell’s bridge which is suitable for low values of Q (i.e. Q < 10 )

SOL 6.26

nodia

Correct option is (C). 4 1 digit display will read from 000.00 to 199.99 So error of 10 counts is equal to 2 = ! 0.10 V For 100 V, the maximum error is

Percentage error SOL 6.27

e = ! (100 # 0.002 + 0.1) = ! 0.3 V = ! 0.3 # 100 % = ! 0.3 % of reading 100

Correct option is (D). Since potential coil is applied across Z2 as shown below

Wattmeter read power consumed by Z2 SOL 6.28

Correct option is (D). Given that full scale current is 5 A


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Chapter 6

Current in shunt Il = IR - I fs = 25 - 5 = 20 A 20 # Rsh = 5 # 0.2 Rsh = 1 = .05 W 20 SOL 6.29

Correct option is (A). Overall gain of the system is 100 = 10 (zero error) g = 1 + 100 b 9 l 100 Gain with error 100 + 10% 110 = 10.091 = g = 9 110 #9 1 + 1 + (100 + 10%) b 100 l 100

nodia error 3 g = 10.091 - 10 - 0.1 100 - 10% 90 Similarly g = = 9 1 + (100 - 10%) 1 + 90 # 9 100 100 = 9.89 error 3 g = 9.89 - 10 -- 0.1 So gain g = 10 ! 0.1 = 10 ! 1%

SOL 6.30

Correct option is (A). At balance condition -j = R2 R 3 wC 4 m - jR 4 wC 4 = R2 R 3 (R + jwL) j R c 4 wC 4 m - jRR 4 wLR 4 + = R2 R 3 R 4 wC 4 wC 4 - jRR 4 LR 4 + = R2 R 3 R 4 wC 4 C4 Comparing real & imaginary parts. RR 4 = R2 R 3 wC 4 wC 4 R = R2 R 3 R4 LR 4 = R R R Similarly, 2 3 4 C4 (R + jwL) c R 4 <

jR2 R 3 wC 4 jR2 R 3 wC 4

L = R2 R3 C 4 SOL 6.31

Correct option is (B). Since Potential coil is connected across the load terminal, so it should be highly resistive, so that all the voltage appears across load.


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Page 375

SOL 6.32

Correct option is (D). A circle is produced when there is a 90c phase difference between vertical and horizontal inputs.

SOL 6.33

Correct option is (C). Wattmeter reading P = VPC ICC VPC " Voltage across potential coil. ICC " Current in current coil. VPC = Vbc = 400+ - 120c ICC = Iac = 400+120c = 4+120c 100 P = 400+ - 120c # 4+120c

Power

nodia = 1600+240c = 1600 # 1 = 800 Watt 2

SOL 6.34

Correct option is (D). Average value of a triangular wave Vav = Vm 3 rms value Vms = Vm 3 Given that Vav = Vm = 10 V 3 So Vrms = Vm = 3 Vav = 10 3 V 3

SOL 6.35

Correct option is (A). Conversion time does not depend on input voltage so it remains same for both type of ADCs.

SOL 6.36


Frequency ratio

meeting points of horizontal tangents fY = fX meeting points of vertical tangents fY =2 4 fX fY = 1 (fX ) 2

w2 = w1 /2 Since the Lissajous figures are ellipse, so there is a phase difference of 90c exists between vertical and horizontal inputs.


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So SOL 6.37

Chapter 6

q (w2 t) = A cos w2 t, w2 = w1 /2

Correct option is (A). Impedance of different branches is given as ZAB = 500 W 1 + 300 W j # 2p # 2 # 103 # 0.398 mF - (- 200j + 300) W = j # 2p # 2 # 103 # 15.91 mH + 300 W - (200j + 300) W

ZBC = ZAD To balance the bridge

ZAB ZCD = ZAD ZBC 500Z = (200j + 300) (- 200j + 300)

nodia 500Z = 130000 Z = (260 + j0) W

SOL 6.38

SOL 6.39

Correct option is (B). Since both the waveform appeared across resistor and inductor are same so the common point is B. Signal Probe S1 is connecte with A, S2 is connected with C and both the grount probes G1 and G2 are connected with common point B. Correct option is (A). To balance the bridge

(R1 + jX1) (R 4 - jX 4) = R2 R 3 (R1 R 4 + X1 X 4) + j (X1 R 4 - R1 X 4) = R2 R 3 comparing real and imaginary parts on both sides of equations ...(1) R 1 R 4 + X1 X 4 = R 2 R 3 ...(2) X 1 R 4 - R 1 X 4 = 0 & X1 = R 1 X4 R4 from eq(1) and (2) it is clear that for balancing the bridge first balance R 4 and then R1 . SOL 6.40

Correct option is (C). From the Calibration pulse we can obtain Voltage (3 V) = 5 = 2.5 V 2 Division Time (3 T) = 1 ms = 1 msec 4 4 Division So amplitude (p-p) of unknown signal is VP - P = 3 V # 5 = 2.5 # 5 = 7.5 V Time period T = 3 T # 8 = 1 # 8 = 2 ms 4

SOL 6.41

Correct option is (A). Reading of wattmeter (Power) in the circuit T Pav = 1 # VIdt = Common are between V - I T 0


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Page 377

total common area = 0 (Positive and negative area are equal) So Pav = 0 SOL 6.42

Correct option is (C). PMMC instrument reads only dc value so

nodia I PMMC =- 8 A rms meter reads rms value so Irms =

=

(- 8) 2 +

(6 2 ) 2 2

64 + 36 = 10 A

Moving iron instrument also reads rms value of current So I MI = 10 mA Reading are (I PMMC, Irms, I MI) = (- 8 A, 10 A, 10 A)

SOL 6.43

Correct option is (D). Given that w =

xy z

log w = log x + log y - log z Maximum error in w % dw w dx x dy y dy y dz z % dw w

So

dy = ! dx ! ! dz x y z = ! 0.5% reading = ! 1% full scale = ! 1 # 100 = ! 1 100 = ! 1 # 100 = ! 5% reading 20 = 1.5% reading = ! 0.5% ! 5% ! 1.5% = ! 7%

SOL 6.44


SOL 6.45



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Chapter 6

In balanced condition there is no current in CD arm so VC = VD Writing node equation at C and D VC - V + VC = 0 & V = V R3 C b R1 + R 3 l R1 R3 V0 - V + VD = 0 & V = V R4 D b R2 + R 4 l R2 R4 So Vb R3 l = Vb R4 l R1 + R 3 R2 + R 4

nodia R 2 R 3 + R 3 R 4 = R1 R 4 + R 3 R 4 R 1 = R 2 R 3 /R 4

SOL 6.46

Correct option is (C). Q-meter works on the principle of series resonance.

At resonance VC = VL and I = V R Quality factor Q = wL = 1 R wCR = wL # I = VL = VC R#I E E Thus, we can obtain Q SOL 6.47

Correct option is (A). PMMC instruments reads DC value only so it reads 2 V.

SOL 6.48

Correct option is (B). Vfs 2 -1 V 14 mv = 3.5 2n - 1 3.5 2n - 1 = 14 # 10- 3 2n - 1 = 250 2n = 251

Resolution of n-bit DAC = So

n


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Page 379

n = 8 bit SOL 6.49

Correct option is (B). We can obtain the frequency ratio as following

meeting points of horizontal tangents fY = fX meeting points of vertical tangents fY =2 4 fX fY = 1 fX 2 There should exist a phase difference(15c) also to produce exact figure of-8.

SOL 6.50

nodia Correct option is (C). The configuration is shown below

It is given that Im = 100 A Range is to be extended to 0 - 500 A, I = 500 A So Im Rm = (I - Im) Rsh 100 # 0.1 = (500 - 100) Rsh Rsh = 100 # 0.1 = 0.025 W 400 SOL 6.51

Correct option is (D). The configuration is shown below

Current in voltmeter is given by IV = E = 180 = .09 A 2000 2000


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I + IV = 2 amp I = 2 - .09 = 1.91 V R = E = 180 = 94.24 W 1.91 I R 0 = 180 = 90 W 2 % error = R - R 0 # 100 = 94.24 - 90 # 100 = 4.71% 90 R0

So

Ideally

SOL 6.52

Chapter 6

Correct option is (A). The measurement system is shown below

nodia Voltmeter reading

1000 (50 kW z 4 MW) 6 MW + 50 kW z 4 MW l = 1000 # .049 = 8.10 V 6 + .049

V =b

SOL 6.53

Correct option is (D). Total power P = P1 + P2 = 10.5 - 2.5 = 8 kW Power factor = cos q Where q = tan- 1 ; 3 b P2 - P1 lE = tan- 1 : 3 # - 13D 8 P2 + P1 =- 70.43c Power factor = cos q = 0.334

SOL 6.54

Correct option is (B). for the dc potentiometer E \ l E1 = l 1 so, E2 l2 l E2 = E1 d 1 n = (1.18) # 680 = 1.34 V 600 l2

SOL 6.55

Correct option is (C). Let the actual voltage and current are I1 and V1 respectively, then


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Page 381

Current in CC is 20 A 20 = I1 b

1000 1000 + 0.02 l

I1 = 20.0004 A - 20 A 200 = V1 - .02 # 20 = 200.40 Power measured Pm = V1 I1 = 20 (200.40) = 4008 W Load power PL = 20 # 200 = 4000 W % Change = Pm - PL = 4008 - 4000 # 100 4000 PL

nodia = 0.2% more

SOL 6.56


We have to obtain n = I I1

I1 = Rsh = 100 = 1 I2 Rm 1000 10 I1 + I 2 = I I1 + 10I1 = I 11I1 = I n = I = 11 I1 SOL 6.57

Correct option is (B). In the following configuration

1 1 = jw C 2p # 100 # 103 # 10 # 10- 12 writing node equation at P Rectance

Xc =


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Chapter 6

VP - 10 + V 1 + 1 - j = 0 Pb 100 100 500 159 l 10 - VP = VP (1.2 - j0.628) 10 = (2.2 - j0.628) VP VP 10 = 4.38 V 2.28 SOL 6.58

Correct option is (A). The torque on the coil is given by t = NIBA N " no. of turns, I " current, B " magnetic field,

nodia So,

SOL 6.59

N = 100 I = 50 mA B = 200 mTA " Area, A = 10 mm # 20 mm

t = 100 # 50 # 10- 3 # 200 # 10- 3 # 200 # 10- 3 # 10- 3 = 200 # 10- 6 Nm

Correct option is (C). Meter constant (A-sec/rev) is given by 14.4 = I speed I 14.4 = K # Power Where ‘K ’ is the meter constant in rev/kWh. I 14.4 = K # VI 15 14.4 = K # 15 # 250 1 K = 250 # 14.4 1 K = 250 # 14.4 b 1000 # 3600 l = 1000 # 3600 = 1000 rev/kWh. 3600

SOL 6.60

Correct option is (B). For moving iron ameter full scale torque is given by tC = 1 I2 dL 2 dq 240 # 10- 6 = 1 (10) 2 dL 2 dq Change in inductance dL = 4.8 mH/radian dq

SOL 6.61

Correct option is (B). In the figure VRY = 415+30c


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Page 383

VBN = 415 +120c 3 Current in current coil IC = VRY = 415+30c Z 100+36.87c = 4.15+ - 6.87

` power factor = 0.8 cos f = 0.8 & f = 36.87c

Power = VI) = 415 +120c # 4.15+6.87c 3 = 994.3+126.87c Reading of wattmeter P = 994.3 ^cos 126.87ch = 994.3 (- 0.60) =- 597 W

SOL 6.62

nodia Correct option is (A). For small values of phase angle IP = nf , f " Phase angle (radians) IS n " turns ratio Magnetizing ampere-turns = 200 So primary current IP = 200 # 1 = 200 amp Turns ratio n = 500 So

Secondary current IS = 5 amp 200 = 500f 5 f (in degrees) = b 180 lb 200 l p 5 # 500 - 4.58c

SOL 6.63

Correct option is (B). Voltage appeared at secondary winding ES = IS # ZL = 5 # 1 = 5 Volts Voltage induced is given by ES = 5= f=

SOL 6.64

2 pfNf , f " flux 2 # 3.14 # 50 # 500 # f 5 = 45 # 10- 6 wb 3 2 # 3.14 # 25 # 10

Correct option is (A). In PMCC instruments, as temperature increases the coil resistance increases. Swamp resistors are connected in series with the moving coil to provide temperature compensation. Swamping resistors is made of magnin, which has a zero-temperature coefficient.


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Chapter 6

SOL 6.65

Correct answer is (). Effect of stray magnetic field is maximum when the operating field and stray fields are parallel.

SOL 6.66


nodia Let

C1 = 300 pF

1 wC 1 R Now when Cx is connected in parallel with variable resistor C1 ' = 200 pF 1 Q = 120 = w (C1 ' + Cx ) R Q = 120 =

So

SOL 6.67

C 1 = C1 ' + C x 300 = 200 + Cx Cx = 100 pF

Correct option is (B). Maximum frequency of input in dual slop A/D converter is given as where

so

Tm = 2n TC fm = 1 " maximum frquency of input Tm fC = 1 " clock frequency TC f fm = Cn , n = 10 2 6 = 10 = 1 kHz (approax) 1024

SOL 6.68

Correct option is (A). Kelvin Double bridge is used for measuring low values of resistances. (P " 2) Low values of capacitances is precisely measured by schering bridge (Q " 3) Inductance of a coil with large time constant or high quality factor is measured by hay’s bridge (R " 5)

SOL 6.69

Correct option is (C). Full scale deflection is produced by a dc current of 1 mA (Idc) fs = 1 mA For full wave reactifier (Idc) fs = 2Im , Im "peak value of ac current p 1 mA = 2Im 3.14


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Page 385

Im = 1.57 mA Full scale ac current (Irms) fs = 1.57 = 1.11 mA 2

V = (Rs + Rm) (Irms) fs 100 = (Rs + 100) (1.11 mA)

nodia 100 = Rs + 100 (1.11 mA) 100 # 900 = Rs + 100 Rs = 89.9 kW

SOL 6.70

Correct option is (B). First the current coil is connected in R-phase and pressure coil is connected between this phase and the neutral as shown below

Reading of wattmeter W1 = IP VP cos q1 , cos q1 = 0.8 & q1 = 36.86c 400 = IL VL cos q1 3 ...(1) 400 = IL VL # 0.8 3 Now when pressure coil is connected between B and Y-phases, the circuit is


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Chapter 6

phasor diagram

nodia Angle Now wattmeter reading From equation (1) so

q2 = 23.14c + 30c = 54.14c2 W2 = VYB IL cos q2 VL IL = 400 # 3 0.8 W2 = 400 # 3 # cos 53.14c 0.8 = 519.5 W

SOL 6.71

Correct option is (C). In a moving-iron ammeter control torque is given as tc = Kq = 1 I2 dL 2 dq Where K " control spring constant q " deflection 2 Given that L = 10 + 3q - q 4 dL = 3 - q mH/rad b 2l dq So,


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Page 387

tc = (25 # 10- 6) q = 1 (5) 2 b 3 - q l # 10- 6 2 2 2q = 3 - q 2 5q = 3 & q = 6 = 1.2 rad. 2 5 SOL 6.72

Correct option is (B). Magnetizing current Im = 250 = 250 amp 1 Primary current I p = 500 amp Secondary current Is = 5 amp I Turn ratio n = p = 500 = 100 5 Is

nodia Total primary current (IT ) =

[primary current (I p)] 2 +

[magnetising current (I m)] 2

IT = =

I p2 + I m2

(500) 2 + (250) 2 = 559.01 amp

Turn ratio n' = IT = 559.01 = 111.80 5 Is Percentage ratio error 3 n = n - nl # 100 nl 100 - 111.80 100 =- 10.55% = # 111.80

SOL 6.73

Correct option is (C). Power read by meter Pm = VI sin (3 - f) Where 3 "Phase angle between supply voltage and pressure coil flux. f "Phase angle of load Here 3 = 85c, f = 60c "a cos f = 0.5 So measured power Pm = 200 # 5 sin (85c - 60c) = 1100 sin 25c = 464.88 W Actual power PO = VI cos f = 220 # 5 # 0.5 = 550 W Error in measurement = Pm - PO = 464.88 - 550 =- 85.12 W For unity power factor cos f = 1 f = 0c So Pm = 220 # 5 sin (85c - 0c) = 1095.81 W Error in Measurement

SOL 6.74

PO = 220 # 5 cos 0c = 1100 = 1095.81 - 1100 =- 4.19 W

Correct option is (A). We can obtain the Lissaju pattern (in X-Y mode) by following method.


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Chapter 6

Vx = Vxm sin wt Vy = Vym sin (wt + 0c) = sin wt Draw Vx and Vy as shown below

For f = 0c,

nodia Divide both Vy and Vx equal parts and match the corresponding points on the screen. Similarly for f = 90c Vx = Vxm sin wt Vy = Vym sin (wt + 90c)

Similarly for f = 3p 2


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Page 389

nodia we can also obtain for 0 < f < 3p 2

***********


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Analog Electronics

Chapter 7

CHAPTER 7 ANALOG ELECTRONICS


ONE MARK

In the Wien Bridge oscillator circuit shown in figure, the bridge is balanced when

nodia 1 (A) R 3 = R1 , w = R 4 R2 R1 C1 R 2 C 2 R C 1 (B) 2 = 2 , w = R1 C1 R1 C1 R 2 C 2 1 (C) R 3 = R1 + C2 , w = R 4 R 2 C1 R1 C1 R 2 C 2 1 (D) R 3 + R1 = C 2 , w = R 4 R 2 C1 R1 C1 R 2 C 2

EE SP 7.2

The magnitude of the mid-band voltage gain of the circuit shown in figure is (assuming h fe of the transistor to be 100)

(A) 1 (B) 10 (C) 20 (D) 100


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Analog Electronics

Page 391

YEAR 2014 EE01

TWO MARKS

EE SP 7.3

Given that the op-amps in the figure are ideal, the output voltage V0 is

EE SP 7.4

nodia (A) ^V1 - V2h (C) ^V1 - V2h /2

(B) 2 ^V1 - V2h (D) ^V1 + V2h

In the figure shown, assume the op-amp to be ideal. Which of the alternatives v i ^w h ? gives the correct Bode plote for the transfer function v i ^w h


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Analog Electronics


Chapter 7

ONE MARK

EE SP 7.5

The transistor in the given circuit should always be in active region. Take VCE^sat h = 0.2 V , VBE = 0.7 V . The maximum value of Rc in W which can be used, is _____.

EE SP 7.6

The sinusoidal ac source in the figure has an rms value of 202 V . Considering all possible values of RL , the minimum value of Rs in W to avoid burnout of the Zener diode is _____.


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Analog Electronics

Page 393


TWO MARKS

Assuming the diodes to be ideal in the figure, for the output to be clipped, the input voltage vi must be outside the range

(A) - 1 V to - 2 V (C) + 1 V to - 2 V

EE SP 7.8

(B) - 2 V to - 4 V (D) + 2 V to - 4 V

nodia

A 10 kHz even-symmetric square ware is passed through a bandpass filter with centre frequency at 30 kHz and 3 dB passband of 6 kHz. The filter output is (A) a highly attenuated square wave at 10 kHz (B) nearly zero (C) a nearly perfect cosine wave at 30 kHz (D) a nearly perfect sine wave at 30 kHz

EE SP 7.9

An oscillator circuit using ideal op-amp and diodes is shown in the figure.

The time duration for + ve part of the cycle is Tt1 and for - ve part is Tt2 . The Tt1 - Tt 2 RC

value of e

will be ______.


ONE MAERK

An operational amplifier circuit is shown in the figure. The output of the circuit for a given input vi is (A) -b R2 l vi (B) -b1 + R2 l vi R1 R1 (C) b1 + R2 l vi (D) + Vsat or - Vsat R1


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Analog Electronics


EE SP 7.11

TWO MARKS

The transfer characteristic of the Op-amp circuit shown in figure is


Chapter 7

ONE MARK

In a voltage-voltage feedback as shown below, which one of the following statements is TRUE if the gain k is increased?


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Analog Electronics

Page 395

(A) The input impedance increases and output impedance decreases (B) The input impedance increases and output impedance also increases (C) The input impedance decreases and output impedance also decreases

nodia (D) The input impedance decreases and output impedance increases

EE SP 7.13

In the circuit shown below what is the output voltage ^Vouth if a silicon transistor Q and an ideal op-amp are used?

(A) - 15 V (C) + 0.7 V

(B) - 0.7 V (D) + 15 V


TWO MARKS

In the circuit shown below the op-amps are ideal. Then, Vout in Volts is

(A) 4

(B) 6


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Analog Electronics

(C) 8 EE SP 7.15

(D) 10

A voltage 1000 sin wt Volts is applied across YZ . Assuming ideal diodes, the voltage measured across WX in Volts, is

nodia (A) sin wt (C) ^sin wt - sin wt h /2

EE SP 7.16

Chapter 7

(B) _sin wt + sin wt i /2 (D) 0 for all t

In the circuit shown below, the knee current of the ideal Zener dioide is 10 mA . To maintain 5 V across RL , the minimum value of RL in W and the minimum power rating of the Zener diode in mW, respectively, are

(A) 125 and 125 (C) 250 and 125

(B) 125 and 250 (D) 250 and 250


ONE MARK

The i -v characteristics of the diode in the circuit given below are v - 0.7 A, v $ 0.7 V i = * 500 0A v < 0. 7 V

The current in the circuit is


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Analog Electronics

Page 397

(A) 10 mA

(B) 9.3 mA

(C) 6.67 mA

(D) 6.2 mA

YEAR 2012

TWO MARKS

EE SP 7.18

The voltage gain Av of the circuit shown below is

EE SP 7.19

nodia (A) Av . 200 (B) Av . 100 (C) Av . 20 (D) Av . 10

The circuit shown is a

1 rad/s (R1 + R2) C (B) high pass filter with f3dB = 1 rad/s R1 C (C) low pass filter with f3dB = 1 rad/s R1 C 1 (D) high pass filter with f3dB = rad/s (R1 + R2) C (A) low pass filter with f3dB =


ONE MARK

A low-pass filter with a cut-off frequency of 30 Hz is cascaded with a high pass filter with a cut-off frequency of 20 Hz. The resultant system of filters will function as (A) an all – pass filter


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Analog Electronics

Chapter 7

(B) an all – stop filter (C) an band stop (band-reject) filter (D) a band – pass filter EE SP 7.21

The CORRECT transfer characteristic is


TWO MARKS

The transistor used in the circuit shown below has a b of 30 and ICBO is negligible


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Analog Electronics

Page 399

If the forward voltage drop of diode is 0.7 V, then the current through collector will be (A) 168 mA (B) 108 mA (C) 20.54 mA (D) 5.36 mA

EE SP 7.23

nodia A clipper circuit is shown below.

Assuming forward voltage drops of the diodes to be 0.7 V, the input-output transfer characteristics of the circuit is


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Analog Electronics

Chapter 7


ONE MARK

Given that the op-amp is ideal, the output voltage vo is

(A) 4 V

nodia (B) 6 V (C) 7.5 V

(D) 12.12 V

EE SP 7.25

Assuming that the diodes in the given circuit are ideal, the voltage V0 is

(A) 4 V (C) 7.5 V

(B) 5 V (D) 12.12 V


TWO MARKS

The transistor circuit shown uses a silicon transistor with VBE = 0.7, IC . IE and a dc current gain of 100. The value of V0 is

(A) 4.65 V (B) 5 V (C) 6.3 V


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Analog Electronics

Page 401

(D) 7.23 V YEAR 2009

ONE MARK

EE SP 7.27

The nature of feedback in the op-amp circuit shown is

EE SP 7.28

nodia (A) Current-Current feedback (B) Voltage-Voltage feedback (C) Current-Voltage feedback (D) Voltage-Current feedback YEAR 2009

TWO MARKS

The following circuit has R = 10 kW, C = 10 mF . The input voltage is a sinusoidal at 50 Hz with an rms value of 10 V. Under ideal conditions, the current Is from the source is

(A) 10p mA leading by 90% (B) 20p mA leading by 90% (C) 10p mA leading by 90% (D) 10p mA lagging by 90% EE SP 7.29

Transformer and emitter follower can both be used for impedance matching at the output of an audio amplifier. The basic relationship between the input power Pin and output power Pout in both the cases is (A) Pin = Pout for both transformer and emitter follower (B) Pin > Pout for both transformer and emitter follower (C) Pin < Pout for transformer and Pin = Pout for emitter follower (D) Pin = Pout for transformer and Pin < Pout for emitter follower


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Analog Electronics

Chapter 7

An ideal op-amp circuit and its input wave form as shown in the figures. The output waveform of this circuit will be


ONE MARK

The equivalent circuits of a diode, during forward biased and reverse biased conditions, are shown in the figure. (1)

(2)

If such a diode is used in clipper circuit of figure given above, the output voltage


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Analog Electronics

Page 403

V0 of the circuit will be

nodia YEAR 2008

EE SP 7.32

TWO MARKS

Two perfectly matched silicon transistor are connected as shown in the figure assuming the b of the transistors to be very high and the forward voltage drop in diodes to be 0.7 V, the value of current I is

(A) 0 mA (C) 4.3 mA EE SP 7.33

(B) 3.6 mA (D) 5.7 mA

In the voltage doubler circuit shown in the figure, the switch ‘S ’ is closed at t = 0 . Assuming diodes D1 and D2 to be ideal, load resistance to be infinite and initial capacitor voltages to be zero. The steady state voltage across capacitor C1 and C2 will be

(A) Vc1 = 10 V,Vc2 = 5 V (B) Vc1 = 10 V,Vc2 =- 5 V (C) Vc1 = 5 V,Vc2 = 10 V (D) Vc1 = 5 V,Vc2 =- 10 V


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Analog Electronics

Chapter 7

The block diagrams of two of half wave rectifiers are shown in the figure. The transfer characteristics of the rectifiers are also shown within the block.

It is desired to make full wave rectifier using above two half-wave rectifiers. The resultants circuit will be

nodia Statement for Linked Answer Questions 35 and 36 A general filter circuit is shown in the figure :


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Analog Electronics

EE SP 7.35

If R1 = R2 = RA and R3 = R4 = RB , the circuit acts as a (A) all pass filter (B) band pass filter (C) high pass filter (D) low pass filter

EE SP 7.36

The output of the this filter is given to the circuit in figure : The gain v/s frequency characteristic of the output (vo) will be

Page 405


ONE MARK

The common emitter forward current gain of the transistor shown is bF = 100

The transistor is operating in (A) Saturation region (B) Cutoff region (C) Reverse active region (D) Forward active region EE SP 7.38

The three-terminal linear voltage regulator is connected to a 10 W load resistor as shown in the figure. If Vin is 10 V, what is the power dissipated in the transistor ?


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EE SP 7.39

Analog Electronics

Chapter 7

(A) 0.6 W

(B) 2.4 W

(C) 4.2 W

(D) 5.4 W

The circuit shown in the figure is

nodia rV R1 < R2 r < R2 (B) a voltage source with voltage V R1 r < R2 V (C) a current source with current c R1 + R2 m r (D) a current source with current c R2 mV R1 + R2 r (A) a voltage source with voltage


TWO MARKS

The input signal Vin shown in the figure is a 1 kHz square wave voltage that alternates between + 7 V and - 7 V with a 50% duty cycle. Both transistor have the same current gain which is large. The circuit delivers power to the load resistor RL . What is the efficiency of this circuit for the given input ? choose the closest answer.

(A) 46% (C) 63%

(B) 55% (D) 92%


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Analog Electronics


Page 407

ONE MARK

For a given sinusoidal input voltage, the voltage waveform at point P of the clamper circuit shown in figure will be

nodia EE SP 7.42

What are the states of the three ideal diodes of the circuit shown in figure ? (A) D1 ON, D2 OFF, D3 OFF (B) D1 OFF, D2 ON, D3 OFF (C) D1 ON, D2 OFF, D3 ON (D) D1 OFF, D2 ON, D3 ON


TWO MARKS

Assuming the diodes D1 and D2 of the circuit shown in figure to be ideal ones, the transfer characteristics of the circuit will be


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EE SP 7.44

Analog Electronics

Chapter 7

nodia

A relaxation oscillator is made using OPAMP as shown in figure. The supply voltages of the OPAMP are !12 V . The voltage waveform at point P will be


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EE SP 7.45

Analog Electronics

Page 409

Consider the circuit shown in figure. If the b of the transistor is 30 and ICBO is 20 mA and the input voltage is + 5 V , the transistor would be operating in

nodia (A) saturation region (B) active region

(C) breakdown region (D) cut-off region

EE SP 7.46

A TTL NOT gate circuit is shown in figure. Assuming VBE = 0.7 V of both the transistors, if Vi = 3.0 V, then the states of the two transistors will be

(A) Q1 ON and Q2 OFF (B) Q1 reverse ON and Q2 OFF (C) Q1 reverse ON and Q2 ON (D) Q1 OFF and Q2 reverse ON YEAR 2005 EE SP 7.47

ONE MARK

Assume that D1 and D2 in figure are ideal diodes. The value of current is


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EE SP 7.48

Analog Electronics

Chapter 7

(A) 0 mA

(B) 0.5 mA

(C) 1 mA

(D) 2 mA

Assume that the N-channel MOSFET shown in the figure is ideal, and that its threshold voltage is + 1.0 V the voltage Vab between nodes a and b is

nodia (A) 5 V (C) 1 V

(B) 2 V (D) 0 V


The common emitter amplifier shown in the figure is biased using a 1 mA ideal current source. The approximate base current value is

(A) 0 mA (C) 100 mA EE SP 7.50

TWO MARKS

(B) 10 mA (D) 1000 mA

The typical frequency response of a two-stage direct coupled voltage amplifier is as shown in figure


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EE SP 7.51

Analog Electronics

nodia

Consider the inverting amplifier, using an ideal operational amplifier shown in the figure. The designer wishes to realize the input resistance seen by the smallsignal source to be as large as possible, while keeping the voltage gain between - 10 and - 25 . The upper limit on RF is 1 MW. The value of R1 should be

(A) Infinity (C) 100 kW

EE SP 7.52

Page 411

(B) 1 MW (D) 40 kW

In the given figure, if the input is a sinusoidal signal, the output will appear as shown


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Analog Electronics

Chapter 7

Common Data For Q. 53 and 54 Assume that the threshold voltage of the N-channel MOSFET shown in figure is + 0.75 V. The output characteristics of the MOSFET are also shown

EE SP 7.53

EE SP 7.54

nodia The transconductance of the MOSFET is (A) 0.75 ms (B) 1 ms (C) 2 ms (D) 10 ms

The voltage gain of the amplifier is (A) + 5 (C) + 10

(B) - 7.5 (D) - 10


The current through the Zener diode in figure is

(A) 33 mA (C) 2 mA EE SP 7.56

ONE MARK

(B) 3.3 mA (D) 0 mA

Two perfectly matched silicon transistor are connected as shown in figure. The value of the current I is


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EE SP 7.57

Analog Electronics

Page 413

(A) 0 mA

(B) 2.3 mA

(C) 4.3 mA

(D) 7.3 mA

nodia The feedback used in the circuit shown in figure can be classified as

(A) shunt-series feedback (C) series-shunt feedback

(B) shunt-shunt feedback (D) series-series feedback


Assuming that the diodes are ideal in figure, the current in diode D1 is

(A) 9 mA (C) 0 mA EE SP 7.59

TWO MARKS

(B) 5 mA (D) - 3 mA

The trans-conductance gm of the transistor shown in figure is 10 mS. The value of the input resistance Rin is


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Analog Electronics

(A) 10.0 kW (C) 5.0 kW

EE SP 7.60

nodia (B) 470 W (D) 1200 W

In the active filter circuit shown in figure, if Q = 1, a pair of poles will be realized with w0 equal to

(A) 1000 rad/s (C) 10 rad/s EE SP 7.62

(B) 8.3 kW (D) 2.5 kW

The value of R for which the PMOS transistor in figure will be biased in linear region is

(A) 220 W (C) 680 W EE SP 7.61

Chapter 7

(B) 100 rad/s (D) 1 rad/s

The input resistance Rin = vx /ix of the circuit in figure is


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Analog Electronics

(A) + 100 kW (C) + 1 MW

Page 415

(B) - 100 kW (D) - 1 MW


ONE MARK

nodia

The variation of drain current with gate-to-source voltage (ID - VGS characteristic) of a MOSFET is shown in figure. The MOSFET is

(A) an n-channel depletion mode device (B) an n-channel enhancement mode device

(C) an p-channel depletion mode device (D) an p-channel enhancement mode device EE SP 7.64

In the circuit of figure, assume that the transistor has hfe = 99 and VBE = 0.7 V. The value of collector current IC of the transistor is approximately

(A) [3.3/3.3] mA (B) [3.3/(3.3+3.3)] mA (C) [3.3/.33] mA (D) [3.3(33+3.3)] mA EE SP 7.65

For the circuit of figure with an ideal operational amplifier, the maximum phase shift of the output vout with reference to the input vin is


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Analog Electronics

(A) 0c (C) + 90c

Chapter 7

(B) - 90c (D) !180c


nodia

For the n-channel enhancement MOSFET shown in figure, the threshold voltage Vth = 2 V. The drain current ID of the MOSFET is 4 mA when the drain resistance RD is 1 kW.If the value of RD is increased to 4 kW, drain current ID will become

(A) 2.8 mA (C) 1.4 mA EE SP 7.67

TWO MARKS

(B) 2.0 mA (D) 1.0 mA

Assuming the operational amplifier to be ideal, the gain vout /vin for the circuit shown in figure is

(A) - 1 (B) - 20 (C) - 100 (D) - 120 EE SP 7.68

A voltage signal 10 sin wt is applied to the circuit with ideal diodes, as shown in figure, The maximum, and minimum values of the output waveform Vout of the circuit are respectively


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Analog Electronics

Page 417

(A) + 10 V and - 10 V (B) + 4 V and - 4 V (C) + 7 V and - 4 V (D) + 4 V and - 7 V

nodia ************


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Analog Electronics

Chapter 7

SOLUTIONS SOL 7.1

Correct option (C). To determine the balancing condition, we reconstruct the Wien bridge as

nodia At balance, we have

z1 z 4 = z 2 z 3

j R2 c R 1 - w C 1 m R 4 = c 1 + j wC 2 R 2 m R 3 R 3 = R 1 + C 2 + j wC R - 1 or b 2 1 wC 1 R 2 l R4 R 2 C1 Comparing real and imaginary parts, we get R 3 = R1 + C 2 R4 R 2 C1 1 and wC 2 R 1 - 1 =0 &w = wC 1 R 2 R1 C1 R 2 C 2

or

SOL 7.2

Correct option (D). We have the transistor circuit as

Its equivalent h -parameter circuit is drawn as


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Analog Electronics

Page 419

For the circuit, we obtain the mid-band gain as Am = Vo Vs h Rl (Rs = 0 as it by passes due to capacitor) = fe Rs + hie 3

= 100 # 10 # 10 = 100 0 + 10 # 103 i.e. the mid-band gain is 100. SOL 7.3

Correct option is (B). We redraw the given op-amp circuit as

nodia Node voltage Vc is connected to inverting mode of op-amp. So, we have the output due to Vc as R Vo+ = f V1 =- R Vc R1 R Again, node voltage Vd is connected to the non-inverting mode of op-amp. So, we have the output due to Vd as R R Vo- =+c1 + f m f V2 = b1 + R l R Vd R1 2R1 R 2R Therefore, the net output of the op-amp is Vo = Vo+ + Vo=- Vc # b R l + Vd R b1 + R l R 2R R =- Vc + Vd # 2 = Vd - Vc 2 Applying KVL in the op-amp circuit, we have

- Vc + IR + I ^2Rh + IR + Vd = 0 or Vd - Vc =- 4IR or Vo =- 4IR Also, we have

- Va + 2IR + Vb = 0 or 2IR = Va - Vb From equations (3) and (4), we get

...(1)

[using equation (1)] ...(2)

...(3)

Vo =- 2 ^Va - Vb h


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Analog Electronics

Chapter 7

...(4) = 2 ^Vb - Va h Since, the voltage at inverting and non-inverting terminals are same for ideal opamp. So, from the op-amp circuit, we have V1 = Vb and V2 = Va Substituting it in equation (4), we obtain Vo = 2 ^V1 - V2h

SOL 7.4

Correct option is (A) We have the op-amp circuit

nodia Let the voltage at inverting terminal of op-amp be X . So, we have X ^s h - Vo ^s h =0 Rf or

X ^s h = Vo ^s h

...(1)

Applying KCL at non-inverting terminal of op-amp, we get X ^s h - Vi ^s h =0 [X ^s h - 0] Cs + R1 Vi ^s h or X ^s h:Cs + 1 D = R1 R1 Vi ^s h or [from equation (1)] Vo ^s h:R1 Cs + 1D = R1 R1 Vo ^s h 1 1 So, = = 1 + R1 Cs 1 + 10-3 Vi ^s h Therefore, the corner frequency for the transfer function is wC = 1-3 = 103 10 Hence, we draw the Bode plot for the function (in decibel), Vo ^wh 1 20 log = 20 log Vi ^wh f 1 + jw3 p 10 The obtained magnitude and phase plots are


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Analog Electronics

Page 421


Given:

VCE^sath = 0.2 VBE = 0.7

nodia b = 100 Writing KVL around the input loop

5 - I B R S - 0. 7 = 0 IB = 5 - 0.7 = 4.3 3 = 2.15 mA 2 kW 2 # 10 IC = bIB = 100 # 2.15 mA = 0.215 A Writing KVL around the output loop

For active region or, So,

SOL 7.6

5 - IC RC - VCE = 0 VCE = 5 - 0.215RC VCE > VCE^sath

VCE > 0.2 V 5 - 0.215RC > 0.2 0.215RC < 5 - 0.2 RC min = 4.8 . 22.32 W 0.215

Correct answer is 300. The given circuit is a bridge full wave rectifier with zener regulator. The voltage across the regulator is a full wave rectified output as shown below.

I S = IZ + I L


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Analog Electronics

Chapter 7

IZ = I S - I L IZ max = IS max - IL min Because when IL will be minimum IZ will be maximum. We have IS max = Vo - VZ = 20 - 5 Rs min Rs min IL min = 0

^1/4h = 1 A IZ max = PZ = 5 20 VZ Substituting these values in eq (1), we get 1 = 20 - 5 20 Rs min

...(1)

(for RL " 3)

Rs min = 15 # 20 = 300 W

SOL 7.7

nodia Correct option is (B). Assume when both the diodes will be OFF.

In that case vo = vi 2 For proper clipping operation, both the diode can not be forward biased together. When both diodes are off, we have vo = vi 2 Let v p1 , vn1 are voltage of anode and cathode of diode D1 and v p2 and vn2 are voltage of anode and cathode of D2 . So vo = v p1 = vi 2 vo = vn1 = vi 2 To make both diodes forward biased v p1 > - 1 V and vn2 < - 2 V vi > - 1 V and vi < - 2 V or 2 2 So clipping operation. SOL 7.8


SOL 7.9

Correct answer is 1.2.

vi =- 2 V to - 4 V range does not allow proper


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Analog Electronics

Page 423

Correct option is (D). We have the op-amp circuit as given below.

nodia For 1. 2. 3.

an operation amplifier, we know that If vi > Vsat ; then output voltage will be Vsat If vi < - Vsat ; then the output voltage will be - Vsat If - Vsat < vi < Vsat ; then the output voltage varies according to the applied input.

For the given op-amp circuit, we assume that

- Vsat < vi < Vsat So, we obtain the output for the corresponding input vi . Applying KCL at noninverting terminal of first op-amp, we have vi - 0 + vi - x = 0 R1 R2 or vi b 1 + 1 l = x R2 R1 R 2 or ...(1) x = b R1 + R 2 l v i R1 where x is the voltage at non-inverting terminal of second op-amp. Again, applying KCL at non-inverting terminal of second op-amp, we get x - 0 + x - v0 = 0 R R v 0 = 2x From eq (1) we have v 0 = 2vi ^R1 + R2h = 2vi b1 + R2 l R1 R1 There is no single option so our assumption is wrong and vi > + Vsat and vi < - Vsat . So, the output voltage is + Vsat and - Vsat . or

SOL 7.11

Correct option is (C). The given op-amp circuit is redrawn below. For the given circuit, we consider the following two cased. Case I : when vi > 0 For this case, the diode conducts. Since, the voltage at negative and positive terminals are same for ideal op-amp. So, for the 1st op-amp, we have V 1- = V 1+ = 0 Therefore, the output of first op-amp is zero, which is applied to negative terminal of second op-amp.


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Analog Electronics

Chapter 7

Thus we get the output vo = 0 .

nodia Case II : when vi < 0 For this case, the diode is open and the output of first op-amp is Vo1 =- R vi =- vi R Therefore, the output of second op-amp is vo =- R vo1 = vi R Thus, we have the output voltage 0 vi > 0 vo = ) vi vi < 0 Hence, the transfer characteristic of op-amp circuit is

SOL 7.12

Correct option is (A). The i/p voltage of the system is given as Vin = V1 + Vf = V1 + k Vout = V1 + k A 0 V1 ^Vout = A 0 V1h = V1 ^1 + k A 0h Therefore, if k is increased then input voltage is also increased so, the input impedance increases. Now, we have Vout = A 0 V1 Vin = A 0 Vin ^1 + k A 0h ^1 + k A 0h Since, Vin is independent of k when seen from output mode, the output voltage decreases with increase in k that leads to the decrease of output impedance. Thus, input impedance increases and output impedance decreases. = A0


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Analog Electronics

Page 425

Correct option is (B). For the given ideal op-amp, negative terminal will be also ground (at zero voltage) and so, the collector terminal of the BJT will be at zero voltage. VC = 0 volt The current in 1 kW resistor is given by I = 5 - 0 = 5 mA 1 kW This current will flow completely through the BJT since, no current will flow into the ideal op-amp ( I/P resistance of ideal op-amp is infinity). So, for BJT we have i.e.,

VC = 0 VB = 0 IC = 5 mA i.e.,the base collector junction is reverse biased (zero voltage) therefore, the collector current (IC ) can have a value only if base-emitter is forward biased. Hence,

nodia & &

SOL 7.14

VBE = 0.7 volts VB - VE = 0.7 0 - Vout = 0.7 & Vout =- 0.7 volt

Correct option is (C). For the given ideal op-Amps we can assume V 2- = V 2+ = V2 (ideal) V 1+ = V 1- = V1 (ideal) So, by voltage division V1 = Vout # 1 2

Vout = 2V1 and, as the I/P current in Op-amp is always zero therefore, there will be no voltage drop across 1 KW in II op-amp i.e., V2 = 1 V V2 - ^- 2h Therefore, V1 - V2 = 1 1 or, or, So,

V1 - 1 = 1 + 2 V1 = 4 Vout = 2V1 = 8 volt


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SOL 7.15

Analog Electronics

Chapter 7

nodia Correct option is (D). Given, the input voltage VYZ = 100 sin wt

For + ve half cycle

VYZ > 0 i.e., VY is a higher voltage than VZ So, the diode will be in cutoff region. Therefore, there will no voltage difference between X and W node. i.e., VWX = 0 Now, for - ve half cycle all the four diodes will active and so, X and W terminal is short circuited i.e., VWX = 0 Hence, VWX = 0 for all t SOL 7.16


From the circuit, we have


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Analog Electronics

Page 427

Is = IZ + I L (1) or, IZ = Is - I L Since, voltage across zener diode is 5 V so, current through 100 W resistor is obtained as Is = 10 - 5 = 0.05 A 100 Therefore, the load current is given by IL = 5 RL Since, for proper operation, we must have IZ $ Iknes So, from Eq. (1), we write 0.05 A - 5 $ 10 mA RL 50 mA - 5 $ 10 mA RL 40 mA $ 5 RL 40 # 10-3 $ 5 RL 1 # RL 5 40 # 10-3 5 # RL 40 # 10-3 or, 125 W # RL Therefore, minimum value of RL = 125 W Now, we know that power rating of Zener diode is given by

nodia PR = VZ IZ^maxh IZ^maxh is maximum current through zener diode in reverse bias. Maximum currrent through zener diode flows when load current is zero. i.e., IZ^maxh = Is = 10 - 5 = 0.05 100 Therefore, SOL 7.17

Correct option is (D). Let v > 0.7 V and diode is forward biased. Applying Kirchoff’s voltage law 10 - i # 1k - v = 0 10 - :v - 0.7 D (1000) - v = 0 500

So, SOL 7.18

PR = 5 # 0.05 W = 250 mW

10 - (v - 0.7) # 2 - v = 0 v = 11.4 = 3.8 V > 0.7 3 i = v - 0.7 = 3.8 - 0.7 = 6.2 mA 500 500

(Assumption is true)

Correct option is (D). DC Analysis :


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Analog Electronics

Chapter 7

Using KVL in input loop, VC - 100IB - 0.7 = 0 VC = 100IB + 0.7 IC - IE = 13.7 - VC = (b + 1) IB 12k 13.7 - VC = 100I B 12 # 103 Solving equation (1) and (2), IB = 0.01 mA Small Signal Analysis : Transforming given input voltage source into equivalent current source.

nodia This is a shunt-shunt feedback amplifier. Given parameters, rp = VT = 25 mV = 2.5 kW IB 0.01 mA b 100 = 0.04 s gm = = rp 2.5 # 1000 Writing KCL at output node v0 + g v + v0 - vp = 0 m p RC RF v 0 : 1 + 1 D + v p :gm - 1 D = 0 RC RF RF Substituting RC = 12 kW, RF = 100 kW, gm = 0.04 s v 0 (9.33 # 10-5) + v p (0.04) = 0 or v 0 =- 428.72Vp Writing KCL at input node vi = v p + v p + v p - vo Rs Rs rp RF = v p : 1 + 1 + 1 D - v 0 = v p (5.1 # 10-4) - v 0 RF Rs rp RF RF Substituting Vp from equation (1) vi = - 5.1 # 10-4 v - v 0 0 428.72 Rs RF vi =- 1.16 # 10-6 v 0 - 1 # 10-5 v 0 Rs = 10 kW 10 # 103

...(1)

...(2)

...(1)


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Analog Electronics

Page 429

vi =- 1.116 # 10-5 10 # 103 1 - 8.96 Av = v 0 = vi 10 # 103 # 1.116 # 10-5 SOL 7.19

Correct option is (B). First we obtain the transfer function.

nodia 0 - Vi (jw) 0 - Vo (jw) + =0 1 R2 jwC + R1 - V (jw) Vo (jw) = 1 i R2 jwC + R1 V (jw) R2 Vo (jw) =- i R1 - j jw1C

1 " 3, so V = 0 o wC 1 " 0, so V (jw) =- R2 V (jw) At w " 3 (higher frequencies), o R1 i wC The filter passes high frequencies so it is a high pass filter. H (jw) = Vo = - R2 Vi R1 - j 1 wC R R 2 2 = H (3) = R1 R1 At w " 0 (Low frequencies),

2 times of maximum gain 6H (3)@ H ^ jw0h = 1 H (3) 2 R2 = 1 b R2 l 1 2 2 R1 R1 + 2 2 w0 C 2R 12 = R 12 + 21 2 & R 12 = 21 2 w0 C w C w0 = 1 R1 C

At 3 dB frequency, gain will be

So,

SOL 7.20



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Analog Electronics

Chapter 7

nodia So, it will act as a Band pass filter.

SOL 7.21

Correct option is (D). Zener Diode is used as stabilizer. The circuit is assumed to be as

We can see that both BE and BC Junction are forwarded biased. So the BJT is operating in saturation. Collector current IC = 12 - 0.2 = 5.36 mA 2.2k Y bIB Note:- In saturation mode IC SOL 7.22



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Analog Electronics

Page 431

The first half of the circuit is a differential amplifier (negative feedback) Va =- (Vi) Second op-amp has a positive feedback, so it acts as an schmitt trigger.

nodia Since

Threshold value

Va =- Vi this is a non-inverting schmitt trigger. VTH = 12 = 6 V 2 VTL =- 6 V

SOL 7.23

Correct option is (C). We can obtain three operating regions depending on whether the Zener and PN diodes are forward biased or reversed biased. 1. vi #- 0.7 V , zener diode becomes forward biased and diode D will be off so the equivalent circuit looks like

The output vo =- 0.7 V 2. When - 0.7 1 vi # 5.7 , both zener and diode D will be off. The circuit is

Output follows input i.e vo = vi Note that zener goes in reverse breakdown(i.e acts as a constant battery) only when difference between its p-n junction voltages exceeds 10 V. 3. When vi > 5.7 V , the diode D will be forward biased and zener remains off, the equivalent circuit is


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Analog Electronics

Chapter 7

vo = 5 + 0.7 = 5.7 V SOL 7.24

Correct option is (B). Since the op-amp is ideal v+ = v- =+ 2 volt Writing node equation v- - 0 + v- - vo = 0 R 2R 2 + (2 - vo) = 0 R 2R

nodia 4 + 2 - vo = 0 or vo = 6 volt

SOL 7.25

Correct option is (B). Given circuit is,

We can observe that diode D2 is always off, whether D1 ,is on or off. So equivalent circuit is.

D1 is ON in this condition and V0 = SOL 7.26

10 10 = 5 volt 10 + 10 #

Correct option is (A). Writing KVL equation for input loop (Base emitter loop) 10 - (10 kW) IB - VBE - V0 = 0 Emitter current IE = V0 100

...(1)

IC - IE = bIB


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Analog Electronics

Page 433

V0 = 100I B 100 V0 IB = 10 # 103

So,

Put IB into equation (1) 10 - (10 # 103)

&

SOL 7.27

V0 - 0.7 - V0 = 0 10 # 103 9.3 - 2V0 = 0 V0 = 9.3 = 4.65 A 2

Correct option is (B). Equivalent circuit of given amplifier

nodia Feedback samples output voltage and adds a negative feedback voltage (vfb) to input. So, it is a voltage-voltage feedback. SOL 7.28

Correct option is (A). Let voltages at positive and negative terminals of op-amp are V+ and Vrespectively, then V+ = V- = Vs (ideal op-amp) In the circuit we have, V- - 0 + V- - V0 (s) = 0 R 1 ` Cs j (RCs) V- + V- - V0 (s) = 0 (1 + RCs) Vs = V0 (s) Similarly current Is is, Is = Vs - V0 R Is = RCs Vs R Is = jwCVs Is = wCVs + + 90% Is = 2pf # 10 # 10 - 6 # 10 Is = 2 # p # 50 # 10 # 10 - 6 # 10 Is = 10p mA, leading by 90%


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Analog Electronics

Chapter 7

Correct option is (D). Input and output power of a transformer is same Pin = Pout For emitter follower, voltage gain (A v) = 1 current gain (Ai) > 1 Power (Pout) = Av Ai Pin Since emitter follower has a high current gain so Pout > Pin

SOL 7.30

Correct option is (D). This is a schmitt trigger circuit, output can takes two states only. VOH =+ 6 volt VOL =- 3 volt Threshold voltages at non-inverting terminals of op-amp is given as VTH - 6 + VTH - 0 = 0 2 1

nodia Similarly

3VTH - 6 = 0 VTH = 2 V (Upper threshold) VTL - (- 3) VTL + =0 2 1

3VTL + 3 = 0 VTL =- 1 V (Lower threshold) Vin < 2 Volt, V0 =+ 6 Volt Vin > 2 Volt, V0 =- 3 Volt Vin < - 1 Volt V0 =+ 6 Volt Vin > - 1 Volt V0 =- 3 Volt

For

Output waveform

SOL 7.31

Correct option is (A). Assume the diode is in reverse bias so equivalent circuit is


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Analog Electronics

Page 435

V0 = 10 sin wt # 10 10 + 10

Output voltage

= 5 sin wt Due to resistor divider, voltage across diode VD < 0 (always). So it in reverse bias for given input.

nodia V0 = 5 sin wt

Output,

SOL 7.32


This is a current mirror circuit. Since b is high so IC1 = IC2, IB1 = IB2 VB = (- 5 + 0.7) =- 4.3 volt Diode D1 is forward biased. So, current I is,

SOL 7.33

I = IC2 = IC1 0 - (- 4.3) = = 4.3 mA 1

Correct option is (D). In positive half cycle of input, diode D1 is in forward bias and D2 is off, the equivalent circuit is

Capacitor C1 will charge upto + 5 volt. VC1 =+ 5 volt In negative halt cycle diode D1 is off and D2 is on.


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Analog Electronics

Chapter 7

Now capacitor VC2 will charge upto - 10 volt in opposite direction. SOL 7.34

Option () is correct. Let input Vin is a sine wave shown below

nodia

According to given transfer characteristics of rectifiers output of rectifier P is.

Similarly output of rectifier Q is

Output of a full wave rectifier is given as

To get output V0 K - gain of op-amp V0 = K (- VP + VQ) So, P should connected at inverting terminal of op-amp and Q with non-inverting terminal.


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Analog Electronics

Page 437

Correct option is (C). w " 0 , so

For low frequencies,

1 "3 wC

Equivalent circuit is,

nodia

Applying node equation at positive and negative input terminals of op-amp. vA - vi + vA - vo = 0 R1 R2 2vA = vi + vo ,

a R1 = R 2 = R A

Similarly,

vA - vi + vA - 0 = 0 R3 R4 2vA = vin ,

a R 3 = R 4 = RB

vo = 0 , iIt will stop low frequency signals. For high frequencies, w " 3 , then 1 " 0 wC Equivalent circuit is, So,

Output, vo = vi So it will pass high frequency signal. This is a high pass filter. SOL 7.36

Correct option is (D). In previous solution cutoff frequency of high pass filter is given by, 1 wh = 2pRA C Here given circuit is a low pass filter with cutoff frequency, 1 2 wL = = 2pRA C 2p R2 C wL = 2wh When both the circuits are connected together, equivalent circuit is, A


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Analog Electronics

Chapter 7

So this is is Band pass filter, amplitude response is given by.

SOL 7.37

Correct option is (D). If transistor is in normal active region, base current can be calculated as following, By applying KVL for input loop,

nodia

10 - IC (1 # 103) - 0.7 - 270 # 103 IB = 0 bIB + 270 IB = 9.3 mA, ` IC = bIB IB (b + 270) = 9.3 mA IB = 9.3 mA = 0.025 mA 270 + 100 In saturation, base current is given by, 10 - IC (1) - VCE - IE (1) = 0 10 = I C (sat) 2

IC (sat) = 5 mA IC (sat) IB(sat) = = 5 = .050 mA b 100 IB 1 IB(sat), so transistor is in forward active region.

SOL 7.38

IC - IE VCE - 0


We can analyze that the transistor is operating in active region. VBE(ON) = 0.6 volt VB - VE = 0.6 6.6 - VE = 0.6 VE = 6.6 - 0.6 = 6 volt At emitter (by applying KCL), IE = IB + IL IE = 6 - 6.6 + 6 - 0.6 amp 1 kW 10 W VCE = VC - VE = 10 - 6 = 4 volt Power dissipated in transistor is given by.


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Analog Electronics

Page 439

PT = VCE # IC = 4 # 0.6 = 2.4 W SOL 7.39

` IC - IE = 0.6 amp

Correct option is (D). This is a voltage-to-current converter circuit. Output current depends on input voltage.

nodia Since op-amp is ideal v+ = v- = v1 Writing node equation. v1 - v + v1 - 0 = 0 R1 R2 v1 c R1 + R2 m = V R1 R1 R2

v1 = V c

R2 R1 + R2 m

Since the op-amp is ideal therefore

iL = i1 = v1 r = V c R2 m r R1 + R2 SOL 7.40

Option () is correct. This is a class-B amplifier whose efficiency is given as h = p VP 4 VCC where VP " peak value of input signal VCC " supply voltage here VP = 7 volt, VCC = 10 volt so,

h = p # 7 # 100 10 4 = 54.95% - 55%

SOL 7.41

Correct option is (A). First we can check for diode D2 . Let diode D2 is OFF then the circuit is


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Analog Electronics

Chapter 7

In the above circuit diode D1 must be ON, as it is connected with 10 V battery now the circuit is

nodia

Because we assumed diode D2 OFF so voltage across it VD2 # 0 and it is possible only when D3 is off.

So, all assumptions are true. SOL 7.42

Correct option is (D). In the positive half cycle of input, Diode D1 will be reverse biased and equivalent circuit is.

Since there is no feed back to the op-amp and op-amp has a high open loop gain so it goes in saturation. Input is applied at inverting terminal so. VP =- VCC =- 12 V In negative half cycle of input, diode D1 is in forward bias and equivalent circuit is shown below.


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Analog Electronics

Page 441

Output VP = Vg + VOp-amp is at virtual ground so V+ = V- = 0 and VP = Vg = 0.7 V Voltage wave form at point P is

SOL 7.43

nodia Correct option is (A). In the circuit when Vi < 10 V, both D1 and D2 are off. So equivalent circuit is,

Output,

Vo = 10 volt

When Vi > 10 V (D1 is in forward bias and D2 is off So the equivalent circuit is,

Output, Vo = Vi Transfer characteristic of the circuit is


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SOL 7.44

Analog Electronics

Chapter 7

Correct option is (C). Here output of the multi vibrator is V0 = ! 12 volt Threshold voltage at positive terminal of op-amp can be obtained as following When output V0 =+ 12 V, equivalent circuit is,

nodia writing node equation at positive terminal of op-amp Vth - 12 + Vth - 0 = 0 10 10 Vth = 6 volt (Positive threshold) So, the capacitor will charge upto 6 volt. When output V0 =- 12 V, the equivalent circuit is.

node equation Vth + 12 + Vth - 0 = 0 2 10 5 Vth + 60 + Vth = 0 Vth =- 10 volt (negative threshold)


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Analog Electronics

Page 443

So the capacitor will discharge upto - 10 volt. At terminal P voltage waveform is.

SOL 7.45

Correct option is (B). Assume that BJT is in active region, thevenin equivalent of input circuit is obtained as

nodia Vth - Vi + Vth - (- 12) = 0 15 100 20Vth - 20Vi + 3Vth + 36 23Vth Vth Thevenin resistance Rth

=0 = 20 # 5 - 36 , Vi = 5 V = 2.78 V = 15 KW || 100 KW = 13.04 KW

So the circuit is

Writing KVL for input loop 2.78 - Rth IB - 0.7 = 0 IB = 0.157 mA Current in saturation is given as, I IB(sat) = C(sat) b


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Analog Electronics

Chapter 7

IC(sat) = 12.2 = 5.4 mA 2.2 IB(sat) = 5.45 mA = 0.181 mA 30

So,

Since IB (sat) > IB , therefore assumption is true. SOL 7.46


SOL 7.47

Correct option is (A). From the circuit we can observe that Diode D1 must be in forward bias (since current is flowing through diode). Let assume that D2 is in reverse bias, so equivalent circuit is.

nodia Voltage Vn is given by

Vn = 1 # 2 = 2 Volt Vp = 0 Vn > Vp (so diode is in reverse bias, assumption is true) Current through D2 is ID2 = 0

SOL 7.48

Correct option is (D). This is a N-channel MOSFET with VGS = 2 V VTH =+ 1 V VDS(sat) = VGS - VTH VDS(sat) = 2 - 1 = 1 V Due to 10 V source VDS > VDS(sat) so the NMOS goes in saturation, channel conductivity is high and a high current flows through drain to source and it acts as a short circuit. So, Vab = 0

SOL 7.49

Correct option is (B). Since the transistor is operating in active region. IE . bIB IB = IE = 1 mA = 10 mA 100 b

SOL 7.50

Correct option is (C). Gain of the inverting amplifier is given by, 6 Av =- RF =- 1 # 10 , R1 R1

RF = 1 MW


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Analog Electronics

R1 =- 1 # 10 Av Av =- 10 to - 25 so value of R1

Page 445

6

6 R1 = 10 = 100 kW 10

for Av =- 10

6 R1' = 10 = 40 kW 25

for Av =- 25

R1 should be as large as possible so R1 = 100 kW SOL 7.51

SOL 7.52

SOL 7.53

SOL 7.54

Correct option is (B). Direct coupled amplifiers or DC-coupled amplifiers provides gain at dc or very low frequency also.

nodia

Correct option is (C). Since there is no feedback in the circuit and ideally op-amp has a very high value of open loop gain, so it goes into saturation (ouput is either + V or - V ) for small values of input. The input is applied to negative terminal of op-amp, so in positive half cycle it saturates to - V and in negative half cycle it goes to + V . Correct option is (B). Trans-conductance of MOSFET is given by gm = 2iD 2VGS (2 - 1) mA = = 1 mS (2 - 1) V

Correct option is (D). Voltage gain can be obtain by small signal equivalent circuit of given amplifier.

vo =- gm vgs RD vgs = vin vo =- gm RD vin Av = vo =- gm RD vi

So, Voltage gain

=- (1 mS) (10 kW) =- 10 SOL 7.55

Correct option is (C). Given circuit,


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Analog Electronics

Chapter 7

In the circuit V1 = 3.5 V (given) Current in zener is. IZ = V1 - VZ RZ = 3.5 - 3.33 = 2 mA 0.1 # 10

SOL 7.56

nodia

Correct option is (C). This is a current mirror circuit. Since VBE is the same in both devices, and transistors are perfectly matched, then IB1 = IB2 and IC1 = IC2 From the circuit we have,

IR = IC1 + IB1 + IB2 = IC1 + 2IB2 = IC2 + 2IC2 b IR = IC2 c1 + 2 m b IR IC2 = I = 2 c1 + b m

a IB1 = IB2 a IC1 = IC2, IC2 = bIB2

IR can be calculate as IR = - 5 + 03.7 =- 4.3 mA 1 # 10 So,

SOL 7.57

I =

4.3 - 4.3 mA 2 1 + ` 100 j

Correct option is (B). The small signal equivalent circuit of given amplifier


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Analog Electronics

Page 447

Here the feedback circuit samples the output voltage and produces a feed back current Ifb which is in shunt with input signal. So this is a shunt-shunt feedback configuration. SOL 7.58

Correct option is (C). From the circuit, we can see the that diode D2 must be in forward Bias. For D1 let assume it is in reverse bias. Voltages at p and n terminal of D1 is given by Vp and Vn Vp < Vn (D1 is reverse biased)

nodia Applying node equation Vp - 5 Vp + 8 + =0 1 1 2Vp =- 3 Vp =- 1.5

Vn = 0 Vp < Vn (so the assumption is true and D1 is in reverse bias) and current in D1 ID1 = 0 mA SOL 7.59

Correct option is (D). The small signal ac equivalent circuit of given amplifier is as following.

Here

Input resistance

RB = (10 kW < 10 kW) = 5 kW gm = 10 ms 50 a gm rp = b & rp = = 5 kW 10 # 10 - 3 Rin = RB < rp = 5 kW < 5 kW = 2.5 kW


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Analog Electronics

Chapter 7

Correct option is (D). For PMOS to be biased in non-saturation region. VSD < VSD(sat) and VSD(sat) = VSG + VT VSD(sat) = 4 - 1

"a VSG = 4 - 0 = 4 volt

= 3 Volt So,

VSD < 3 VS - VD < 3 4 - ID R < 3 1 < ID R ID R > 1, R > 1000 W

SOL 7.61

SOL 7.62

nodia

ID = 1 mA

Option () is correct.

Option (B is correct. If op-amp is ideal, no current will enter in op-amp. So current ix is v - vy ...(1) ix = x 1 # 106 (ideal op-amp) v+ = v- = vx vx - vy + vx - 0 3 = 0 3 100 # 10 10 # 10 vx - vy + 10vx = 0 11vx = vy

...(2)

For equation (1) & (2) ix = vx - 11v6 x =- 10v6x 1 # 10 10 Input impedance of the circuit. 6 Rin = vx =- 10 =- 100 kW 10 ix SOL 7.63

Correct option is (C). Since there exists a drain current for zero gate voltage (VGS = 0), so it is a depletion mode device. ID increases for negative values of gate voltages so it is a p-type depletion mode device.

SOL 7.64

Correct option is (B). Applying KVL in input loop, 4 - (33 # 103) IB - VBE - (3.3 # 103) IE = 0 4 - (33 # 103) IB - 0.7 - (3.3 # 103) (hfe + 1) IB = 0

a IE = (hfe + 1) IB

3.3 = 6(33 # 103) + (3.3 # 103) (99 + 1)@IB 3.3 IB = 3 33 # 10 + 3.3 # 103 # 100


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Analog Electronics

Page 449

IC = hfe IB = SOL 7.65

3.3 99 # 3.3 mA mA = 0.33 + 3.3 [0.33 + 3.3] # 100

Correct option is (D). Let the voltages at positive and negative terminals of op-amp are v+ and vrespectively. Then by applying nodal equations. v- - vin + v- - vout = 0 R1 R1 2 v-- = vin + vout

..(1)

Similarly, v+ - vin v -0 + + =0 R 1 c jwC m

nodia v+ - vin + v+ (jwCR) = 0 v+ (1 + jwCR) = Vin By equation (1) and (2) we have 2vin = vin + vout 1 + jwCR 2 - 1E = vout vin ; 1 + jwCR (1 - jwCR) vout = vin 1 + jwCR

..(2)

"a v+ = v- (ideal op-amp)

Phase shift in output is given by q = tan - 1 (- wCR) - tan - 1 (wCR) = p - tan - 1 (wCR) - tan - 1 (wCR) = p - 2 tan - 1 (wCR) Maximum phase shift SOL 7.66

q =p

Correct option is (A). For a n -channel enhancement mode MOSFET transition point is given by, a VTH = 2 volt VDS (sat) = VGS - VTH VDS (sat) = VGS - 2 From the circuit, VDS = VGS So VDS (sat) = VDS - 2 & VDS = VDS (sat) + 2 VDS > VDS (sat) Therefore transistor is in saturation region and current equation is given by. ID = K (VGS - VTH ) 2 4 = K (VGS - 2) 2 VGS is given by VGS = VDS = 10 - ID RD = 10 - 4 # 1 = 6 Volt 4 = K (6 - 2) 2 K =1 4 ' ' Now RD is increased to 4 kW, Let current is ID' and voltages are VDS = VGS So,


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Analog Electronics

Chapter 7

Applying current equation. ' - VTH ) 2 ID' = K (VGS ' - 2) 2 ID' = 1 (VGS 4 ' ' = 10 - ID' # RD' = 10 - 4ID' = VDS VGS

4ID' = (10 - 4ID' - 2) 2 = (8 - 4ID' ) 2

So,

= 16 (2 - ID' ) 2 ID' = 4 (4 + I'D2 - 4ID' ) 4I'D2 - 17 + 16 = 0 I'D2 = 2.84 mA SOL 7.67

Correct option is (D). Let the voltages at input terminals of op-amp are v- and v+ respectively. So, v+ = v- = 0 (ideal op-amp)

nodia Applying node equation at negative terminal of op-amp, 0 - vin + 0 - vx = 0 1 10 vx - 0 + vx - vout + vx - 0 = 0 At node x 10 10 1

From equation (1),

SOL 7.68

...(1)

vx + vx - vout + 10vx = 0 12 vx = vout vx = vout 12 vin + vx = 0 1 10 vin =- vout 120 vout =- 120 vin

Correct option is (D). In the positive half cycle (when Vin > 4 V ) diode D2 conducts and D1 will be off so the equivalent circuit is,


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Analog Electronics

Page 451

Vout = + 4 Volt In the negative half cycle diode D1 conducts and D2 will be off so the circuit is,

Applying KVL Vin - 10I + 4 - 10I = 0 Vin + 4 = I 20 Vin =- 10 V (Maximum value in negative half cycle) So, I = - 10 + 4 =- 3 mA 20 10 Vin - Vout = I 10 - 10 - Vout =- 3 10 10

nodia Vout =- (10 - 3) Vout =- 7 volt ***********


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Digital Electronics

Chapter 8

CHAPTER 8 DIGITAL ELECTRONICS


ONE MARK

A cascade of three identical modulo-5 counters has an overall modulus of (A) 5 (B) 25 (C) 125

(D) 625


EE SP 8.2

TWO MARKS

An output device is interfaced with 8-bit microprocessor 8085A. The interfacing circuit is shown in figure

The interfacing circuit makes use of 3 Line to 8 Line decoder having 3 enables lines E1 , E 2 , E 3 . The address of the device is (A) 50 H (B) 5000 H (C) A0 H (D) A000 H EE SP 8.3

Which of the following logic circuits is a realization of the function F whose Karnaugh map is shown in figure


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Digital Electronics


EE SP 8.4

ONE MARK

Which of the following is an invalid state in an 8-4-2-1 Binary Coded Decimal counter (A) 1 0 0 0 (B) 1 0 0 1 (C) 0 0 1 1 (D) 1 1 0 0 YEAR 2014 EE02

EE SP 8.5

EE SP 8.6

Page 453

TWO MARKS

The SOP (sum of products) form of a Boolean function is S ^0, 1, 3, 7, 11h, where inputs are A, B , C , D (A is MSB, and D is LSB). The equivalent minimized expression of the function is (A) ^B + C hÂ + C hÂ + B h^C + D h (B) ^B + C hÂ + C hÂ + C h^C + D h (C) ^B + C hÂ + C hÂ + C h^C + D h (D) ^B + C hÂ + B hÂ + B h^C + D h A JK flip flop can be implemented by T flip-flops. Identify the correct implementation.


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Digital Electronics

Chapter 8

nodia EE SP 8.7

In an 8085 microprocessor, the following program is executed Address location Instruction 2000H XRA A 2001H MVI B, 04H 2003H MVI A, 03H 2005H RAR 2006H DCR B 2007H JNZ 2005 200AH HLT At the end of program, register A contains (A) 60H (B) 30H (C) 06H (D) 03H


ONE MAKR

A state diagram of a logic gate which exhibits a delay in the output is shown in the figure, where X is the don’t care condition, and Q is the output representing the state.


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Digital Electronics

Page 455

The logic gate represented by the state diagram is (A) XOR (B) OR (C) AND (D) NAND EE SP 8.9

nodia

In 8085A microprocessor, the operation performed by the instruction LHLD 2100 H is (A) ^Hh ! 21 H , ^Lh ! 00 H (B) ^Hh ! M ^2100hH , ^Lh ! M ^2101 Hh (C) ^Hh ! M ^2101 Hh, ^Lh ! M ^2100 Hh (D) ^Hh ! 00 H , ^Lh ! 21 H YEAR 2014 EE03

EE SP 8.10

TWO MARKS

Two monoshot multivibrators, one positive edge triggered ^M1h and another negative edge triggered ^M2h, are connected as shown in figure

The monoshots M1 and M2 when triggered produce pulses of width T1 and T2 respectively, where T1 > T2 . The steady state output voltage v 0 of the circuit is


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EE SP 8.11

Digital Electronics

Chapter 8

nodia

A 3-bit gray counter is used to control the output of the multiplexer as shown in the figure. The initial state of the counter is 000 2 . The output is pulled high. The output of the circuit follows the sequence

(A) I 0 , 1, 1, I1 , I 3 , 1, 1, I2 (B) I 0 , 1, I1 , 1, I2 , 1, I 3 , 1 (C) 1, I 0 , 1, I1 , I2 , 1, I 3 , 1 (D) I 0 , I1 , I2 , I 3 , I 0 , I1 , I2 , I 3 YEAR 2013 EE SP 8.12

ONE MARK

A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles (A) and AND gate


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Digital Electronics

Page 457

(B) an OR gate (C) an XOR gate (D) a NAND gate YEAR 2013 EE SP 8.13

In the circuit shown below, Q1 has negligible collector-to-emitter saturation voltage and the diode drops negligible voltage across it under forward bias. If Vcc is + 5 V , X and Y are digital signals with 0 V as logic 0 and Vcc as logic 1, then the Boolean expression for Z is

nodia (A) XY (C) XY

EE SP 8.14

TWO MARKS

(B) XY (D) XY

The clock frequency applied to the digital circuit shown in the figure below is 1 kHz. If the initial state of the output of the flip-flop is 0, then the frequency of the output waveform Q in kHz is

(A) 0.25 (B) 0.5 (C) 1 (D) 2 YEAR 2012 EE SP 8.15

ONE MARK

In the sum of products function f (X, Y, Z) = are (A) XY, XY (B) XY, X Y Z , XY Z (C) XY Z , XYZ, XY

/ (2, 3, 4, 5), the prime implicants


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Digital Electronics

Chapter 8

(D) XY Z , XYZ, XY Z , XY Z EE SP 8.16

The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B . The number of combinations for which the output is logic 1, is (A) 4 (B) 6 (C) 8

EE SP 8.17

(D) 10

Consider the given circuit

nodia In this circuit, the race around (A) does not occur

(B) occur when CLK = 0 (C) occur when CLK = 1 and A = B = 1 (D) occur when CLK = 1 and A = B = 0 YEAR 2012

EE SP 8.18

TWO MARKS

The state transition diagram for the logic circuit shown is


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Digital Electronics

Page 459


ONE MARK

The output Y of the logic circuit given below is

(A) 1

(B) 0

(C) X

(D) X


TWO MARKS

nodia A two bit counter circuit is shown below

It the state QA QB of the counter at the clock time tn is ‘10’ then the state QA QB of the counter at tn + 3 (after three clock cycles) will be (A) 00 (B) 01 (C) 10

EE SP 8.21

(D) 11

A portion of the main program to call a subroutine SUB in an 8085 environment is given below. h LXI D, DISP LP : CALL SUB LP+3 h It is desired that control be returned to LP+DISP+3 when the RET instruction is executed in the subroutine. The set of instructions that precede the RET instruction in the subroutine are

POP D (A) DAD H PUSH D

POP DAD INX (B) INX INX PUSH

POP H (C) DAD D PUSH H

XTHL INX D (D) INX D INX D XTHL

H D H H H H


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Digital Electronics

Chapter 8


TWO MARKS

The TTL circuit shown in the figure is fed with the waveform X (also shown). All gates have equal propagation delay of 10 ns. The output Y of the circuit is

nodia Statement For Linked Answer Questions: 23 and 24 The following Karnaugh map represents a function F .

EE SP 8.23

EE SP 8.24

A minimized form of the function F is (A) F = X Y + YZ (C) F = X Y + Y Z

(B) F = X Y + YZ (D) F = X Y + Y Z

Which of the following circuits is a realization of the above function F ?


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EE SP 8.25

Digital Electronics

Page 461

nodia

When a “CALL Addr” instruction is executed, the CPU carries out the following sequential operations internally : Note: (R) means content of register R ((R)) means content of memory location pointed to by R. PC means Program Counter SP means Stack Pointer (A) (SP) incremented (B) (PC)!Addr (PC)!Addr ((SP))!(PC)

(C) (PC)!Addr (SP) incremented ((SP))!(PC)

((SP))!(PC)

(SP) incremented

(D)

((SP))!(PC) (SP) incremented (PC)!Addr


ONE MARK

The following circuit has a source voltage VS as shown in the graph. The current through the circuit is also shown.

The element connected between a and b could be


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EE SP 8.27

Digital Electronics

Chapter 8

The increasing order of speed of data access for the following device is (I) Cache Memory (II) CD-ROM (III) Dynamic RAM (IV) Processor Registers

EE SP 8.28

(V) Magnetic Tape (A) (V), (II), (III), (IV), (I)

(B) (V), (II), (III), (I), (IV)

(C) (II), (I), (III), (IV), (V)

(D) (V), (II), (I), (III), (IV)

nodia

The complete set of only those Logic Gates designated as Universal Gates is (A) NOT, OR and AND Gates (B) XNOR, NOR and NAND Gates (C) NOR and NAND Gates (D) XOR, NOR and NAND Gates YEAR 2009

EE SP 8.29

TWO MARKS

In an 8085 microprocessor, the contents of the Accumulator, after the following instructions are executed will become XRA A MVI B, F0 H SUB B (A) 01 H (B) 0F H (C) F0 H (D) 10 H YEAR 2008

EE SP 8.30

TWO MARKS

A waveform generator circuit using OPAMPs is shown in the figure. It produces a triangular wave at point ‘P’ with a peak to peak voltage of 5 V for Vi = 0 V .

If the voltage Vi is made + 2.5 V, the voltage waveform at point ‘P’ will become


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EE SP 8.31

Digital Electronics

Page 463

nodia

A 3-line to 8-line decoder, with active low outputs, is used to implement a 3-variable Boolean function as shown in the figure

The simplified form of Boolean function F (A, B, C) implemented in ‘Product of Sum’ form will be (A) (X + Z) (X + Y + Z ) (Y + Z) (B) (X + Z ) (X + Y + Z) (Y + Z ) (C) (X + Y + Z) (X + Y + Z) (X + Y + Z) (X + Y + Z ) (D) (X + Y + Z) (X + Y + Z ) (X + Y + Z) (X + Y + Z ) EE SP 8.32

The content of some of the memory location in an 8085 accumulator based system are given below Address

Content

g

g

26FE

00

26FF

01

2700

02

2701

03

2702

04

g

g

The content of stack (SP), program counter (PC) and (H,L) are 2700 H, 2100 H and 0000 H respectively. When the following sequence of instruction are executed. 2100 H: DAD SP


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Digital Electronics

Chapter 8

2101 H: PCHL the content of (SP) and (PC) at the end of execution will be (A) PC = 2102 H, SP = 2700 H (B) PC = 2700 H, SP = 2700 H (C) PC = 2800 H, SP = 26FE H (D) PC = 2A02 H, SP = 2702 H YEAR 2007 EE SP 8.33

ONE MARK

A, B, C and D are input, and Y is the output bit in the XOR gate circuit of the figure below. Which of the following statements about the sum S of A, B, C, D and Y is correct ?

nodia (A) S is always with zero or odd (B) S is always either zero or even (C) S = 1 only if the sum of A, B, C and D is even (D) S = 1 only if the sum of A, B, C and D is odd YEAR 2007 EE SP 8.34

TWO MARKS

The switch S in the circuit of the figure is initially closed, it is opened at time t = 0 . You may neglect the zener diode forward voltage drops. What is the behavior of vout for t > 0 ?

(A) It makes a transition from - 5 V (B) It makes a transition from - 5 V (C) It makes a transition from + 5 V (D) It makes a transition from + 5 V EE SP 8.35

to + 5 V at t = 12.98 ms to + 5 V at t = 2.57 ms to - 5 V at t = 12.98 ms to - 5 V at t = 2.57 ms

IC 555 in the adjacent figure is configured as an astable multi-vibrator. It is enabled to to oscillate at t = 0 by applying a high input to pin 4. The pin


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Digital Electronics

Page 465

description is : 1 and 8-supply; 2-trigger; 4-reset; 6-threshold 7-discharge. The waveform appearing across the capacitor starting from t = 0 , as observed on a storage CRO is

nodia EE SP 8.36

The Octal equivalent of HEX and number AB.CD is (A) 253.314 (B) 253.632 (C) 526.314 (D) 526.632 YEAR 2006

EE SP 8.37

TWO MARKS

A student has made a 3-bit binary down counter and connected to the R-2R ladder type DAC, [Gain = (- 1 kW/2R)] as shown in figure to generate a staircase waveform. The output achieved is different as shown in figure. What could be the possible cause of this error ?


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Digital Electronics

Chapter 8

(A) The resistance values are incorrect option. (B) The counter is not working properly (C) The connection from the counter of DAC is not proper (D) The R and 2R resistance are interchanged EE SP 8.38

nodia

A 4 # 1 MUX is used to implement a 3-input Boolean function as shown in figure. The Boolean function F (A, B, C) implemented is

(A) F (A, B, C) = S (1, 2, 4, 6) (B) F (A, B, C) = S (1, 2, 6) (C) F (A, B, C) = S (2, 4, 5, 6) (D) F (A, B, C) = S (1, 5, 6)

EE SP 8.39

A software delay subroutine is written as given below : DELAY : MVI H, 255D MVI L, 255D LOOP : DCR L JNZ LOOP DCR H JNZ LOOP How many times DCR L instruction will be executed ? (A) 255 (B) 510 (C) 65025 (D) 65279

EE SP 8.40

In an 8085 A microprocessor based system, it is desired to increment the contents of memory location whose address is available in (D,E) register pair and store the result in same location. The sequence of instruction is (A) XCHG (B) XCHG INR M INX H (C) INX D (D) INR M XCHG XCHG


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Digital Electronics

Page 467

YEAR 2005

ONE MARK

EE SP 8.41

The 8085 assembly language instruction that stores the content of H and L register into the memory locations 2050H and 2051H , respectively is (A) SPHL 2050H (B) SPHL 2051H (C) SHLD 2050H (D) STAX 2050H

EE SP 8.42

The digital circuit shown in the figure works as

nodia (A) JK flip-flop

(B) Clocked RS flip-flop

(C) T flip-flop

(D) Ring counter

YEAR 2005

EE SP 8.43

EE SP 8.44

TWO MARKS

Select the circuit which will produce the given output Q for the input signals X1 and X2 given in the figure

If X1 and X2 are the inputs to the circuit shown in the figure, the output Q is


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Digital Electronics

(A) X1 + X2 (C) X1 : X2 EE SP 8.45

Chapter 8

(B) X1 : X2 (D) X1 : X2

In the figure, as long as X1 = 1 and X2 = 1, the output Q remains

nodia (A) at 1

(B) at 0

(C) at its initial value

(D) unstable

YEAR 2004

EE SP 8.46

The digital circuit using two inverters shown in figure will act as

(A) a bistable multi-vibrator (C) a monostable multi-vibrator EE SP 8.47

(B) an astable multi-vibrator (D) an oscillator

The voltage comparator shown in figure can be used in the analog-to-digital conversion as

(A) a 1-bit quantizer (C) a 4-bit quantizer

(B) a 2-bit quantizer (D) a 8-bit quantizer


ONE MARK

TWO MARKS

The simplified form of the Boolean expression Y = (A $ BC + D) (A $ D + B $ C ) can be written as


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Digital Electronics

Page 469

(A) A $ D + B $ C $ D (B) AD + B $ C $ D (C) (A + D) (B $ C + D ) (D) A $ D + BC $ D EE SP 8.49

EE SP 8.50

EE SP 8.51

If the following program is executed in a microprocessor, the number of instruction cycle it will take from START to HALT is START MVI A, 14H ; Move 14 H to register A SHIFT RLC ; Rotate left without carry JNZ SHIFT ; Jump on non-zero to SHIFT HALT (A) 4 (B) 8 (C) 13 (D) 16

nodia

The digital circuit shown in figure generates a modified clock pulse at the output. Choose the correct output waveform from the options given below.

A digit circuit which compares two numbers A3 A2 A1 A0 and B 3 B2 B1 B 0 is shown in figure. To get output Y = 0 , choose one pair of correct input numbers.


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EE SP 8.52

Digital Electronics

Chapter 8

(A) 1010, 1010

(B) 0101, 0101

(C) 0010, 0010

(D) 1010, 1011

nodia

In the Schmitt trigger circuit shown in figure, if VCE (sat) = 0.1 V , the output logic low level (VOL) is

(A) 1.25 V (C) 2.50 V

(B) 1.35 V (D) 5.00 V


ONE MARK

Figure shows a 4 to 1 MUX to be used to implement the sum S of a 1-bit full adder with input bits P and Q and the carry input Cin . Which of the following combinations of inputs to I0, I1, I2 and I3 of the MUX will realize the sum S ?

(A) I0 = I1 = Cin; I2 = I3 = Cin (B) I0 = I1 = C in; I2 = I3 = Cin (C) I0 = I3 = Cin; I1 = I2 = Cin (D) I0 = I3 = C in; I1 = I2 = Cin EE SP 8.54

When a program is being executed in an 8085 microprocessor, its Program


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Digital Electronics

Page 471

Counter contains (A) the number of instructions in the current program that have already been executed (B) the total number of instructions in the program being executed. (C) the memory address of the instruction that is being currently executed (D) the memory address of the instruction that is to be executed next YEAR 2003 EE SP 8.55

TWO MARKS

The circuit of figure shows a 555 Timer IC connected as an astable multi-vibrator. The value of the capacitor C is 10 nF. The values of the resistors RA and RB for a frequency of 10 kHz and a duty cycle of 0.75 for the output voltage waveform are

nodia (A) RA = 3.62 kW, RB = 3.62 kW (B) RA = 3.62 kW, RB = 7.25 kW (C) RA = 7.25 kW, RB = 3.62 kW (D) RA = 7.25 kW, RB = 7.25 kW EE SP 8.56

The boolean expression X Y Z + XY Z + XYZ + XY Z + XYZ can be simplified to (A) XZ + X Z + YZ (B) XY + Y Z + YZ (C) XY + YZ + XZ

EE SP 8.57

(D) XY + YZ + X Z

The shift register shown in figure is initially loaded with the bit pattern 1010. Subsequently the shift register is clocked, and with each clock pulse the pattern gets shifted by one bit position to the right. With each shift, the bit at the serial input is pushed to the left most position (msb). After how many clock pulses will the content of the shift register become 1010 again ?

(A) 3 (C) 11

(B) 7 (D) 15


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Digital Electronics

An X-Y flip-flop, whose Characteristic Table is given below is to be implemented using a J-K flip flop

(A) J = X, K = Y (C) J = Y, K = X

EE SP 8.59

EE SP 8.60

Chapter 8

(B) J = X, K = Y (D) J = Y , K = X

nodia

A memory system has a total of 8 memory chips each with 12 address lines and 4 data lines, The total size of the memory system is (A) 16 kbytes (B) 32 kbytes (C) 48 kbytes (D) 64 kbytes The following program is written for an 8085 microprocessor to add two bytes located at memory addresses 1FFE and 1FFF LXI H, 1FFE MOV B, M INR L MOV A, M ADD B INR L MOV M, A XOR A On completion of the execution of the program, the result of addition is found (A) in the register A (B) at the memory address 1000 (C) at the memory address 1F00 (D) at the memory address 2000 ************


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Digital Electronics

Page 473

SOLUTIONS SOL 8.1

Correct option (C). The cascaded connection of three identical modulo-5 counters is shown below.

Since, overall modulus of cascade of n modulus-5 counters is given by Overall modulas = 5 # 5 # .....n times Hence, for the given system, we obtain Overall modulas = 5 # 5 # 5 (n = 3 ) = 125

SOL 8.2

nodia Correct option is (B). Given the interfacing circuit,

As the output port is at 2 ^010h. Hence, the input to the interfacing circuit is I2 I1 I 0 = 010 or A15 A14 A13 = 010 Now, for E1 to be enable, we have or A12 A11 = 10 Therefore, we have the value at address lines as A 15 A 14 A 13 A 12 A 11 0 1 0 1 0 By default at starting the other address lines A10 A 9 .....A 0 should be zero. Thus, we have the overall port address as A 15 A 14 A 13 A 12 A 11 A 10 A 9 A 8 A 7 A 6 A 5 A 4 A 3 A 2 A 1 A 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 which is equivalent to 5000 H SOL 8.3



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Digital Electronics

Chapter 8

Using minimization technique, we redraw the given Karnaugh map as

For the given Karnaugh map, the output function is Y = C A + CB This function is realized by the logic circuit shown below.

nodia SOL 8.4

Correct option is (D). 8-4-2-1 Binary coded decimal counter valid as long as the value is less than 9 or equal to 9. (A) 1000 = 8 # 9 (B) 1001 = 9 # 9 (C) 0011 = 3 # 9 (D) 1100 = 12 > 9 Invalid

SOL 8.5

Correct option is (A). SOP Form, F = S ^0, 1, 3, 7, 11h POS Form, F = P ^2, 4, 5, 6, 8, 9, 10, 12, 13, 14, 15h Now we represent the POS form on the k-map


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Digital Electronics

Page 475

Here we enter 0’s in those cells correspond to the max terms present in the given POS function. Fmin = ^B + C hÂ + C hÂ + B h^C + D h

So,

SOL 8.6

nodia

Correct option is (B) Here, T flip-flop is available and we want to convert it into J -K flip-flop. Hence J and K are the external input and T is the actual input to the existing flip-flop. So, we have to find expression of T in terms of J , K and Qn . Step 1: Construct the present-state-and-next-state table of a J -K flip-flop as shown in Table 7.12.21. Step 2: We write the values of input T that is required to change the state of the flip-flop from Qn to Qn + 1 . The complete conversion table is as shown in Table 7.12.22. 7.12.22: Conversion Table External Inputs

Present State

Next State

Flip-flop Input

J

K

Qn

Qn + 1

T

0

0

0

0

0

0

0

1

1

0

0

1

0

0

0

0

1

1

0

1

1

0

0

1

1

1

0

1

1

0

1

1

0

1

1

1

1

1

0

1

Step 3: Draw K-map for T

Writing expression of T in POS form


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Digital Electronics

Chapter 8

T = ^J + q n h^K + q n h Thus the logic diagram showing the conversion of T flip-flop to J -K flip-flop is show below.

SOL 8.7

Correct option is (A). XRA A " A ! 00H Ex-OR (operation with accumulate) MVI B, 04H 2001H B ! 04H (04H started in B) 2003H MVI A, 03H A ! 03H (03H started in A)

nodia 2005H carry 2006H 2007H So

RAR

DCR B JNZ 2005

B 03H 02H 01H 00H SOL 8.8

rotate accumulater right with

C 1 1 0 0

B ! 03H (B decremented by one) Jump to 2005H till B not comes equal to zero

A 01H 80H COH 60H

Correct option is (D). We have the state diagram of a logic gate,

For the state diagram, we form the truth table (irrespective of the current state) as A

B

X

0

0

1

0

1

1

1

0

1

1

1

0

Thus, from the truth table, it is clear that the logic gate is NAND gate


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Digital Electronics

Page 477

Correct option is (C). For the 8085A microprocessor, given instruction is LHLD 2100 H This operation loads registers register L and H with the content in memory at locations 2100 H and 2101 H respectively, i.e.

^Hh ! M ^2101hH and ^Lh ! M ^2100hH SOL 8.10

Correct option is (C). We have the multivibrators circuit as shown below

nodia

When monoshot multivibrator M1 is triggered, it produces a pulse of T1 duration.

Similarly, when monoshot multivibrator M2 is triggered, it produces a pulse of T2 duration

Let, the output Q2 is initially high. So, the output vo will be high for time period T2 .

For this period, Q2 = 0 So, the output of AND gate is low. Since, M1 is positive triggered. So, M1 will be OFF for duration T2 . After T2 time, Q2 = 0


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Digital Electronics

Chapter 8

and Q2 = 1 So, the output of AND gate is 1; i.e. M1 is triggered. Hence, output Q1 is high for duration T1 . Since, M2 is negative edge triggered, So M2 will be off for the duration T1 and hence, output vo will be as

After T1 time, the output Q1 becomes low, and the M2 gets ON (due to negative edge triggering). Hence, the output vo again goes high for duration T2 . Thus, the complete output waveform is

SOL 8.11

nodia Correct option is (A). We have the given logic circuit

Initial state of gray counter is ^000h2 . So, we have the outputs A 0 = 0 and A1 = 0 and A2 = 0 For the enable input to multiplexer, we have If E = 0 , then it will select any one of I 0 , I1 , I2 , I 3 . If E = 1, then it will be in high impedance state. So, we get the output of multiplexer for the different inputs (output of 3-bit gray counter) as Initial counter output 000 I0 001 1 (High impedance state) 011 1 (High impedance state) 010 I1 110 I3 111 1 (High impedance state) 101 1 (High impedance state)


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Digital Electronics

100 SOL 8.12

Page 479

I2

Correct option is (C). Let A denotes the position of switch at ground floor and B denotes the position of switch at upper floor. The switch can be either in up position or down position. Following are the truth table given for different combinations of A and B A

B

Y(Bulb)

up(1)

up(1)

OFF(0)

Down(0)

Down(0)

OFF(0)

up(1)

Down(0)

ON(1)

Down(0)

up(1)

ON(1)

nodia

When the switches A and B are both up or both down, output will be zero (i.e. Bulb will be OFF). Any of the switch changes its position leads to the ON state of bulb. Hence, from the truth table, we get Y = A5B i.e., the XOR gate.

SOL 8.13

Correct option is (B). For the given circuit, we can make the truth table as below X Y Z 0 0 0 0 1 1 1 0 0 1 1 0

Logic 0 means voltage is v = 0 volt and logic 1 means voltage is 5 volt For x = 0 , y = 0 , Transistor is at cut off mode and diode is forward biased. Since, there is no drop across forward biased diode. So, Z =Y=0 For x = 0 , y = 1, Again Transistor is in cutoff mode, and diode is forward biased. with no current flowing through resistor. So, Z =Y=1 For x = 1, y = 0 , Transistor is in saturation mode and so, z directly connected to ground irrespective of any value of Y . i.e., Z = 0 (ground) Similarly for X = Y = 1 Z = 0 (ground) Hence, from the obtained truth table, we get Z =XY SOL 8.14

Correct option is (B). From the given logic diagram, we obtain X = ^Q 5 Q h $ ^Q 5 Q h = 0 = 1


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Digital Electronics

Chapter 8

So, the input is always ‘1’ at T , since, clock is - ve edge trigged therefore, at the negative edge Q changes its state as shown in waveform below

nodia

Hence, as obtained from the waveform, time period of Q is double to that of CLK I/p and so, frequency is 12 of clock frequency Thus, fQ = FCLK = 1 = 0.5 kHz 2 2

SOL 8.15

Correct option is (A). Prime implicants are the terms that we get by solving K-map

F = XY + XY 1prime 44 2 44 3 implicants SOL 8.16


Y = 1, when A > B A = a1 a 0, B = b1 b 0 a1

a0

b1

b0

Y

0

1

0

0

1

1

0

0

0

1

1

0

0

1

1

1

1

0

0

1

1

1

0

1

1

1

1

1

0

1

Total combination = 6


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Digital Electronics

Page 481

Correct option is (A). The given circuit is

Condition for the race-around It occurs when the output of the circuit (Y1, Y2) oscillates between ‘0’ and ‘1’ checking it from the options. 1. Option (A): When CLK = 0 Output of the NAND gate will be A1 = B1 = 0 = 1. Due to these input to the next NAND gate, Y2 = Y1 : 1 = Y1 and Y1 = Y2 : 1 = Y2 . If Y1 = 0 , Y2 = Y1 = 1 and it will remain the same and doesn’t oscillate. If Y2 = 0 , Y1 = Y2 = 1 and it will also remain the same for the clock period. So, it won’t oscillate for CLK = 0 . So, here race around doesn’t occur for the condition CLK = 0 . 2. Option (C): When CLK = 1, A = B = 1 A1 = B1 = 0 and so Y1 = Y2 = 1 And it will remain same for the clock period. So race around doesn’t occur for the condition. 3. Option (D): When CLK = 1, A = B = 0 So, A1 = B1 = 1 And again as described for Option (B) race around doesn’t occur for the condition. So, Option (A) will be correct.

nodia SOL 8.18

Correct option is (D). Let Qn + 1 is next state and Qn is the present state. From the given below figure. D = Y = AX 0 + AX1 Qn + 1 = D = AX 0 + AX1 If A = 0, If A = 1, So state diagram is

SOL 8.19

Qn + 1 = A Qn + AQn Qn + 1 = Qn Qn + 1 = Qn

X 0 = Q , X1 = Q (toggle of previous state)



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Digital Electronics

Chapter 8

Y = X 5 X = X X + XX = XX + X X = X + X = 1 SOL 8.20

Correct option is (C). The characteristics equation of the JK flip-flop is Q n + 1 = JQ n + KQn From figure it is clear that J = QB ; K = QB

Qn + 1 is the next state

The output of JK flip flop QA (n + 1) = QB QA + QB QA = QB (QA + QA) = QB Output of T flip-flop QB (n + 1) = Q A

SOL 8.21

SOL 8.22

nodia Clock pulse

QA

QB

QA (n + 1)

QB (n + 1)

Initially(tn )

1

0

1

0

tn + 1

1

0

1

0

tn + 2

1

0

1

0

tn + 3

1

0

1

0

Correct option is (C). LXI D, DISP LP : CALL SUB LP + 3 When CALL SUB is executed LP+3 value is pushed(inserted) in the stack. POP H & HL = LP + 3 DAD D & HL = HL + DE = LP + 3 + DE PUSH H & The last two value of the stack will be HL value i.e, LP + DISP + 3 Correct option is (A). The circuit is

Output Y is written as Y = X 5 B Since each gate has a propagation delay of 10 ns.


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SOL 8.23

Digital Electronics

Page 483

nodia

Correct option is (B). Function F can be minimized by grouping of all 1’s in K-map as following.

F = X Y + YZ

SOL 8.24

SOL 8.25

Correct option is (D). Since F = X Y + YZ In option (D)

Correct option is (D). CALL, Address performs two operations (1) PUSH PC &Save the contents of PC (Program Counter) into stack. SP = SP - 2 (decrement) ((SP)) ! (PC) (2) Addr stored in PC. (PC) ! Addr

SOL 8.26

Correct option is (A). Figure shows current characteristic of diode during switching.

SOL 8.27

Correct option is (B). The increasing order of speed is as following


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Digital Electronics

Chapter 8

Magnetic tape> CD-ROM> Dynamic RAM>Cache Memory>Processor register SOL 8.28

Correct option is (). NOR and NAND gates considered as universal gates.

SOL 8.29

Correct option is (D). For the given instruction set, XRA A & XOR A with A & A = 0 MVI B, F0 H&B = F0 H SUB B &A = A - B A = 00000000 B = 1111 0 0 0 0

nodia 2’s complement of (- B) = 0 0 0 1 0 0 0 0 A + (- B) = A - B = 0 0 0 1 0 0 0 0 = 10 H

SOL 8.30

SOL 8.31

Correct option is ()

Correct option is (B). In SOP form, F is written as F = Sm (1, 3, 5, 6) = X Y Z + X YZ + XY Z + XYZ Solving from K- map

F = X Z + Y Z + XYZ In POS form F = (Y + Z) (X + Z) (X + Y + Z ) Since all outputs are active low so each input in above expression is complemented F = (Y + Z ) (X + Z ) (X + Y + Z) SOL 8.32

Correct option is (B). Given that

SP = 2700 H PC = 2100 H HL = 0000 H Executing given instruction set in following steps, DAD SP & Add register pair (SP) to HL register HL = HL + SP HL = 0000 H + 2700 H HL = 2700 H PCHL & Load program counter with HL contents PC = HL = 2700 H So after execution contents are,


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Digital Electronics

Page 485

PC = 2700 H, HL = 2700 H SOL 8.33

Correct option is (D). In the circuit output Y is given as Y = [A 5 B] 5 [C 5 D] Output Y will be 1 if no. of 1’s in the input is odd.

SOL 8.34

Correct option is (B). In the circuit the capacitor starts charging from 0 V (as switch was initially closed) towards a steady state value of 20 V. for t < 0 (initial) for t " 3 (steady state)

nodia

So at any time t , voltage across capacitor (i.e. at inverting terminal of op-amp) is given by vc (t) = vc (3) + [vc (0) - vc (3)] e

-t RC

-t RC

vc (t) = 20 (1 - e ) Voltage at positive terminal of op-amp v+ - vout v+ - 0 + =0 100 10 v+ = 10 vout 11 Due to zener diodes, - 5 # vout # + 5 So, v+ = 10 (5) V 11 Transistor form - 5 V to + 5 V occurs when capacitor charges upto v+ . So 20 (1 - e - t/RC ) = 10 # 5 11 1 - e - t/RC = 5 22 17 = e - t/RC 22 t = RC ln ` 22 j = 1 # 103 # .01 # 10 - 6 # 0.257 = 2.57 msec 17 Voltage waveforms in the circuit is shown below


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SOL 8.35

Digital Electronics

Chapter 8

Correct option is (B). In a 555 astable multi vibrator circuit, charging of capacitor occurs through resistor (RA + RB) and discharging through resistor RB only. Time for charging and discharging is given as.

nodia

TC = 0.693 (RA + RB) C = 0.693 RB C But in the given circuit the diode will go in the forward bias during charging, so the capacitor will charge through resistor RA only and discharge through RB only. RA = RB TC = TD

a So

SOL 8.36

Correct option is (B). First convert the given number from hexadecimal to its binary equivalent, then binary to octal. Hexadecimal no. AB. CD 1 0 10 S 1 0 1 1 $ 1A BB 1 0B0C S 11 0 1 Binary equivalent S C D A B To convert in octal group three binary digits together as shown 010 S 101 S 0 1 1 $S 11 0 S 0 11 S 010 S 5 2 3 6 3 2 (AB.CD) H = (253.632) 8

So, SOL 8.37

Correct option is ()

SOL 8.38

Correct option is (A). Function F can be obtain as, F = I0 S1 S0 + I1 S1 S0 + I2 S1 S0 + I3 S1 S0 = AB C + A B C + 1 $ BC + 0 $ BC = AB C + A BC + BC = AB C + A BC + BC (A + A) = AB C + A BC + ABC + A BC = S (1, 2, 4, 6)

SOL 8.39

Correct option is (A). MVI H and MVI L stores the value 255 in H and L registers. DCR L decrements L by 1 and JNZ checks whether the value of L is zero or not. So DCR L executed 255 times till value of L becomes ‘0’.


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Digital Electronics

Page 487

Then DCR H will be executed and it goes to ‘Loop’ again, since L is of 8 bit so no more decrement possible and it terminates. SOL 8.40

SOL 8.41

SOL 8.42

Correct option is (A). XCHG & Exchange the contain of DE register pair with HL pair So now addresses of memory locations are stored in HL pair. INR M & Increment the contents of memory whose address is stored in HL pair. Correct option is (C). SHLD transfers contain of HL pair to memory location. & L " M [2050H] SHLD 2050 H " M [2051H]

nodia Correct option is (C). Let the present state is Q(t), so input to D-flip flop is given by, D = Q (t) 5 X Next state can be obtained as, Q (t + 1) = D

= Q (t) 5 X

and

= Q (t) X + Q (t) X = Q (t), if X = 1 Q (t + 1) = Q (t), if X = 0

So the circuit behaves as a T flip flop. SOL 8.43

Correct option is (B). From the given input output waveforms truth table for the circuit is drawn as X1 X2 Q 1 0 1 0 0 1 0 1 0 In option (A), for X1 = 1, Q = 0 so it is eliminated. In option (C), for X1 = 0, Q = 0 (always), so it is also eliminated. In option (D), for X1 = 0, Q = 1, which does not match the truth table. Only option (B) satisfies the truth table.

SOL 8.44

Correct option is (D). In the given circuit NMOS Q1 and Q3 makes an inverter circuit. Q4 and Q5 are in parallel works as an OR circuit and Q2 is an output inverter. So output is Q = X1 + X2 = X1 .X2

SOL 8.45

Correct option is (D). Let Q (t) is the present state then from the circuit,


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Digital Electronics

Chapter 8

So, the next state is given by Q (t + 1) = Q (t) (unstable) SOL 8.46

SOL 8.47

Correct option is (A). In the given circuit output is stable for both 1 or 0. So it is a bistable multivibrator.

nodia

Correct option is (A). Since there are two levels (+ VCC or - VCC ) of output in the given comparator circuit. For an n -bit Quantizer 2n = No. of levels 2n = 2 n =1

SOL 8.48

Correct option is (A). Given Boolean expression,

Y = (A $ BC + D) (A $ D + B $ C ) = (A $ BCD) + (ABC $ B $ C ) + (AD) + B C D = A BCD + AD + B C D = AD (BC + 1) + B C D = AD + B C D SOL 8.49

Correct option is (C). The program is executed in following steps. START MVI A, 14H " one instruction cycle. RLC & rotate accumulator left without carry RLC is executed 6 times till value of accumulator becomes zero. JNZ, JNZ checks whether accumulator value is zero or not, it is executed 5 times. HALT " 1-instruction cycle.


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Digital Electronics

Page 489

nodia So total no. of instruction cycles are n = 1+6+5+1 = 13 SOL 8.50

Correct option is (B). In the given circuit, waveforms are given as,

SOL 8.51

Correct option is (D). In the given circuit, output is given as. Y = (A0 5 B0) 9 (A1 5 B1) 9 (A2 5 B2) 9 (A3 5 B3) For option (A) Y = (1 5 1) 9 (0 5 0) 9 (1 5 1) 9 (0 5 0) = 0909090 =1


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Digital Electronics

Chapter 8

For option (B) Y = (0 5 0) 9 (1 5 1) 9 (0 5 0) 9 (1 5 1) = 0909090 =1 For option (C) Y = (0 5 0) 9 (0 5 0) 9 (1 5 1) 9 (0 5 0) = 0909090 =1 For option (D) Y = (1 5 1) 9 (0 5 0) 9 (1 5 1) 9 (0 5 1) = 0909091 =0

SOL 8.52

nodia Correct option is (B). In the given circuit Vi = 0 V So, transistor Q1 is in cut-off region and Q2 is in saturation. 5 - IC RC - VCE(sat) - 1.25 = 0 5 - IC RC - 0.1 - 1.25 = 0

5 - IC RC = 1.35 V0 = 1.35

SOL 8.53

V0 = 5 - IC R

Correct option is (C). In given circuit MUX implements a 1-bit full adder, so output of MUX is given by. F = Sum = A 5 Q 5 Cin Truth table can be obtain as. P

Q

Cin

Sum

0

0

0

0

0

0

1

1

0

1

0

1


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Digital Electronics

0

1

1

0

1

0

0

1

1

0

1

0

1

1

0

0

1

1

1

1

Page 491

Sum = P Q Cin + PQ Cin + P Q Cin + P Q Cin Output of MUX can be written as F = P Q $ I0 + PQ $ I1 + PQ $ I2 + PQ $ I3 I0 = Cin, I1 = Cin, I2 = Cin, I3 = Cin

Inputs are, SOL 8.54

SOL 8.55

nodia

Correct option is (D). Program counter contains address of the instruction that is to be executed next. Correct option is (C). In the circuit, the capacitor charges through resistor (RA + RB) and discharges through RB . Charging and discharging time is given as.

Frequency

TC = 0.693 (RA + RB) C TD = 0.693 RB C 1 1 f= 1 = = T TD + TC 0.693 (RA + 2RB) C 1 = 10 # 103 -9 0.693 (RA + 2RB) # 10 # 10

14.4 # 103 = RA + 2RB duty cycle = TC = 0.75 T 0.693 (RA + RB) C =3 4 0.693 (RA + 2RB) C 4RA + 4RB = 3RA + 6RB RA = 2RB

...(2)

2RA = 14.4 # 103 RA = 7.21 kW RB = 3.60 kW

From (1) and (2) and SOL 8.56

...(1)

Correct option is (B). Given boolean expression can be written as, F = XYZ + X Y Z + XY Z + XYZ + XYZ = X YZ + Y Z (X + X ) + XY (Z + Z) = XYZ + Y Z + XY = Y Z + Y (X + X Z ) a A + BC = (A + B) (A + C) = Y Z + Y (X + X ) (X + Z ) = Y Z + Y (X + Z ) = Y Z + YX + YZ


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Digital Electronics

Chapter 8


X = X1 5 X 0 , Y = X 2 Serial Input Z = X 5 Y = [X1 5 X0] 5 X2 Truth table for the circuit can be obtain as.

nodia Clock pulse

Serial Input

Shift register

Initially

1

1010

1

0

1101

2

0

0110

3

0

0011

4

1

0001

5

0

1000

6

1

0100

7

1

1010

So after 7 clock pulses contents of the shift register is 1010 again. SOL 8.58

Correct option is (D). Characteristic table of the X-Y flip flop is obtained as. X

Y

Qn

Qn+1

0

0

0

1

0

0

1

1

0

1

0

0

0

1

1

1

1

0

0

1

1

0

1

0

1

1

0

0

1

1

0

0

Solving from k-map


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Digital Electronics

Page 493

Characteristic equation of X-Y flip flop is Qn + 1 = Y Qn + XQn Characteristic equation of a J-K flip-flop is given by Qn + 1 = KQn + J Qn by comparing above two characteristic equations J =Y, K=X

SOL 8.59

nodia Correct option is (A). Total size of the memory system is given by.

= (212 # 4) # 8 bits

= 214 # 8 bits = 214 Bytes = 16 K bytes

SOL 8.60

Correct option is (C). Executing all the instructions one by one.

LXI H, 1FFE & H = (1F) H, L = (FE) H MOV B, M & B = Memory [HL] = Memory [1FFE]

INR L & L = L + (1) H = (FF) H MOV A, M & A = Memory [HL] = Memory [1FFF] ADD B & A = A + B INR L & L = L + (1) H = (FF) H + (1) H = 00 MOV M, A & Memory [HL] = A Memory [1F00] = A XOR A & A = A XOR A = 0 So the result of addition is stored at memory address 1F00. ***********


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Power Electronics

Chapter 9

CHAPTER 9 POWER ELECTRONICS


ONE MARK

The figure shows the circuit of a rectifier fed from a 230 V (rms), 50 Hz sinusoidal voltage source. If we want to replace the current source with a resistor so that the rms value of the current supplied by the voltage source remains unchanged, the value of the resistance (in ohms) is _____. (Assume diodes to be ideal.)

nodia EE SP 9.2

Figure shows four electronic switches (i), (ii), (iii) and (iv). Which of the switches can block voltages of either polarity (applied between terminals ‘a’ and ‘b’) when the active device is in the OFF state ?

(A) (i), (ii) and (iii) (C) (ii) and (iii)

(B) (ii), (iii) and (iv) (D) (i) and (iv)


TWO MARKS

The figure shows the circuit diagram of a rectifier. The load consists of a resistance 10 W and an inductance 0.05 H connected in series. Assuming ideal thyristor and ideal diode, the thyristor firing angle (in degree) needed to obtain an average load voltage of 70 V is ____.


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EE SP 9.4

Power Electronics

Page 495

Figure (i) shows the circuit diagram of a chopper. The switch S in the circuit in figure (i) is switched such that the voltage vD across the diode has the wave shape as shown in figure (ii). The capacitance C is large so that the voltage across it is constant. If switch S and the diode are ideal, the peak to peak ripple (in A) in the inductor current is ______

nodia EE SP 9.5

The figure shows one period of the output voltage of an inverter a should be chosen such that 60c < a < 90c. If rms value of the fundamental component is 50 V, then a in degree is _____.


ONE MARK

A step-up chopper is used to feed a load at 400 V dc from a 250 V dc source. The inductor current is continuous. If the ‘off’ time of the switch is 20 ms , the


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Power Electronics

Chapter 9

switching frequency of the chopper in kHz is _____. YEAR 2014 EE02 EE SP 9.7

EE SP 9.8

TWO MARKS

A fully controlled converter bridge feeds a highly inductive load with ripple free load current. The input supply (vs ) to the bridge is a sinusoidal source. Triggering angle of the bridge converter is a = 30c. The input power factor of the bridge is

nodia

A single-phase SCR based ac regulator is feeding power to a load consisting of 5 W resistance and 16 mH inductance. The input supply is 230 V, 50 Hz ac. The maximum firing angle at which the voltage across the device becomes zero all throughout and the rms value of current through SCR, under this operating condition, are (A) 30c and 46 A (B) 30c and 23 A (C) 45c and 23 A

EE SP 9.9

(D) 45c and 32 A

The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of 50 ms is applied to the SCR. The maximum value of R in W to ensure successful firing of the SCR is ______.


TWO MARKS

A hysteresis type TTL inverter is used to realize an oscillator in the circuit shown in the figure.


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Power Electronics

Page 497

If the lower and upper trigger level voltages are 0.9 V and 1.7 V, the period (in ms), for which output is LOW, is _____. EE SP 9.11

A three-phase fully controlled bridge converter is fed through star-delta transformer as shown in the figure.

nodia

The converter is operated at a firing angle of 30c. Assuming the load current Î 0h to be virtually constant at 1 p.u. and transformer to be an ideal one, the input phase current waveform is


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Power Electronics

Chapter 9

n EE SP 9.12

EE SP 9.13

A diode circuit feeds an ideal inductor as shown in the figure. Given vs = 100 sin ^wt h V , where w = 100p rad/s , and L = 31.83 mH . The initial value of inductor current is zero. Switch S is closed at t = 2.5 ms . The peak value of inductor current iL (in A) in the first cycle is _____.

nodia

A single-phase voltage source inverter shown in figure is feeding power to a load. The triggering pulses of the devices are also shown in the figure.

If the load current is sinusoidal and is zero at 0, p , 2p ....., the node voltage VAO has the waveform


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Power Electronics

Page 499


TWO MARKS

Thyristor T in the figure below is initially off and is triggered with a single pulse of width 10 ms . It is given that L = b 100 l mH and C = b 100 l mF . Assuming p p latching and holding currents of the thyristor are both zero and the initial charge on C is zero, T conducts for


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Power Electronics

(A) 10 ms (C) 100 ms EE SP 9.15

Chapter 9

(B) 50 ms (D) 200 ms

The separately excited dc motor in the figure below has a rated armature current of 20 A and a rated armature voltage of 150 V. An ideal chopper switching at 5 kHz is used to control the armature voltage. If La = 0.1 mH , Ra = 1 W , neglecting armature reaction, the duty ratio of the chopper to obtain 50% of the rated torque at the rated speed and the rated field current is

nodia (A) 0.4

(B) 0.5

(C) 0.6

(D) 0.7


In the figure shown below, the chopper feeds a resistive load from a battery source. MOSFET Q is switched at 250 kHz, with duty ratio of 0.4. All elements of the circuit are assumed to be ideal

EE SP 9.16

The Peak to Peak source current ripple in amps is (A) 0.96 (B) 0.144 (C) 0.192 (D) 0.228

EE SP 9.17

The average source current in Amps in steady-state is (A) 3/2 (B) 5/3 (C) 5/2 (D) 15/4

Statement for Linked Answer Questions: 18 and 19 The Voltage Source Inverter (VSI) shown in the figure below is switched to provide a 50 Hz, square wave ac output voltage Vo across an RL load. Reference polarity of Vo and reference direction of the output current io are indicated in the figure. It is given that R = 3 ohms, L = 9.55 mH .


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EE SP 9.18

EE SP 9.19

Power Electronics

Page 501

In the interval when V0 < 0 and i 0 > 0 the pair of devices which conducts the load current is (B) Q 3, Q 4 (A) Q1, Q2 (C) D1, D2 (D) D 3, D 4

nodia

Appropriate transition i.e., Zero Voltage Switching ^ZVS h/Zero Current Switching ^ZCS h of the IGBTs during turn-on/turn-off is (A) ZVS during turn off (B) ZVS during turn-on (C) ZCS during turn off (D) ZCS during turn-on YEAR 2012

ONE MARK

EE SP 9.20

A half-controlled single-phase bridge rectifier is supplying an R-L load. It is operated at a firing angle a and the load current is continuous. The fraction of cycle that the freewheeling diode conducts is (A) 1/2 (B) (1 - a/p) (C) a/2p (D) a/p

EE SP 9.21

The typical ratio of latching current to holding current in a 20 A thyristor is (A) 5.0 (B) 2.0 (C) 1.0 (D) 0.5 YEAR 2012

EE SP 9.22

TWO MARKS

In the circuit shown, an ideal switch S is operated at 100 kHz with a duty ratio of 50%. Given that Dic is 1.6 A peak-to-peak and I 0 is 5 A dc, the peak current in S , is


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Power Electronics

(A) 6.6 A (C) 5.8 A

Chapter 9

(B) 5.0 A (D) 4.2 A

Common Data for Questions 23 and 24 In the 3-phase inverter circuit shown, the load is balanced and the gating scheme is 180c conduction mode. All the switching devices are ideal.

EE SP 9.23

EE SP 9.24

nodia The rms value of load phase voltage is (A) 106.1 V (C) 212.2 V

(B) 141.4 V (D) 282.8 V

If the dc bus voltage Vd = 300 V, the power consumed by 3-phase load is (A) 1.5 kW (B) 2.0 kW (C) 2.5 kW (D) 3.0 kW YEAR 2011

EE SP 9.25

ONE MARK

A three phase current source inverter used for the speed control of an induction motor is to be realized using MOSFET switches as shown below. Switches S1 to S6 are identical switches.

The proper configuration for realizing switches S1 to S6 is


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EE SP 9.26

Power Electronics

Page 503

nodia Circuit turn-off time of an SCR is defined as the time (A) taken by the SCR turn to be off (B) required for the SCR current to become zero (C) for which the SCR is reverse biased by the commutation circuit

(D) for which the SCR is reverse biased to reduce its current below the holding current YEAR 2011 EE SP 9.27

TWO MARKS

A voltage commutated chopper circuit, operated at 500 Hz, is shown below.

If the maximum value of load current is 10 A, then the maximum current through the main (M) and auxiliary (A) thyristors will be (A) iM max = 12 A and iA max = 10 A (B) iM max = 12 A and iA max = 2 A (C) iM max = 10 A and iA max = 12 A (D) iM max = 10 A and iA max = 8 A


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Power Electronics

Chapter 9

Statement for Linked Answer Questions: 28 and 29 A solar energy installation utilize a three – phase bridge converter to feed energy into power system through a transformer of 400 V/400 V, as shown below.

The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance 10 W. EE SP 9.28

EE SP 9.29

The maximum current through the battery will be (A) 14 A (B) 40 A (C) 80 A (D) 94 A

nodia The kVA rating of the input transformer is (A) 53.2 kVA (B) 46.0 kVA (C) 22.6 kVA

(D) 7.5 kVA

YEAR 2010

EE SP 9.30

ONE MARK

The power electronic converter shown in the figure has a single-pole doublethrow switch. The pole P of the switch is connected alternately to throws A and B. The converter shown is a

(A) step down chopper (buck converter) (B) half-wave rectifier (C) step-up chopper (boost converter) (D) full-wave rectifier EE SP 9.31

Figure shows a composite switch consisting of a power transistor (BJT) in series with a diode. Assuming that the transistor switch and the diode are ideal, the I -V characteristic of the composite switch is


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EE SP 9.32

Power Electronics

Page 505

nodia

The fully controlled thyristor converter in the figure is fed from a single-phase source. When the firing angle is 0c, the dc output voltage of the converter is 300 V. What will be the output voltage for a firing angle of 60c, assuming continuous conduction

(A) 150 V (B) 210 V (C) 300 V (D) 100p V YEAR 2009 EE SP 9.33

ONE MARK

An SCR is considered to be a semi-controlled device because (A) It can be turned OFF but not ON with a gate pulse. (B) It conducts only during one half-cycle of an alternating current wave. (C) It can be turned ON but not OFF with a gate pulse. (D) It can be turned ON only during one half-cycle of an alternating voltage wave. YEAR 2009

EE SP 9.34

TWO MARKS

The circuit shows an ideal diode connected to a pure inductor and is connected to a purely sinusoidal 50 Hz voltage source. Under ideal conditions the current waveform through the inductor will look like.


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EE SP 9.35

Power Electronics

nodia

The Current Source Inverter shown in figure is operated by alternately turning on thyristor pairs (T1, T2) and (T3, T4). If the load is purely resistive, the theoretical maximum output frequency obtainable will be

(A) 125 kHz (C) 500 kHz EE SP 9.36

Chapter 9

(B) 250 kHz (D) 50 kHz

In the chopper circuit shown, the main thyristor (TM) is operated at a duty ratio of 0.8 which is much larger the commutation interval. If the maximum allowable reapplied dv/dt on TM is 50 V/ ms , what should be the theoretical minimum value of C1 ? Assume current ripple through L 0 to be negligible.

(A) 0.2 mF

(B) 0.02 mF


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Power Electronics

(C) 2 mF EE SP 9.37

Page 507

(D) 20 mF

Match the switch arrangements on the top row to the steady-state V -I characteristics on the lower row. The steady state operating points are shown by large black dots.

nodia (A) P-I, Q-II, R-III, S-IV (C) P-IV, Q-III, R-I, S-II

(B) P-II, Q-IV, R-I, S-III (D) P-IV, Q-III, R-II, S-I

YEAR 2008

EE SP 9.38

ONE MARK

In the single phase voltage controller circuit shown in the figure, for what range of triggering angle (a), the input voltage (V0) is not controllable ?

(A) 0c < a < 45c (C) 90c < a < 180c EE SP 9.39

(B) 45c < a < 135c (D) 135c < a < 180c

A 3-phase voltage source inverter is operated in 180c conduction mode. Which one of the following statements is true ? (A) Both pole-voltage and line-voltage will have 3rd harmonic components (B) Pole-voltage will have 3rd harmonic component but line-voltage will be free from 3rd harmonic (C) Line-voltage will have 3rd harmonic component but pole-voltage will be free from 3rd harmonic (D) Both pole-voltage and line-voltage will be free from 3rd harmonic components


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Power Electronics


Chapter 9

TWO MARKS

The truth table of monoshot shown in the figure is given in the table below :

Two monoshots, one positive edge triggered and other negative edge triggered, are connected shown in the figure, The pulse widths of the two monoshot outputs Q1 and Q2 are TON and TON respectively. 1

2

nodia

The frequency and the duty cycle of the signal at Q1 will respectively be 1 (A) f = , D= 1 5 TON + TON TON 1 (B) f = , D= TON + TON TON + TON TON (C) f = 1 , D = TON TON + TON TON (D) f = 1 , D = TON TON + TON 1

2

1

2

2

1

2

1

1

1

2

1

2

1

2

EE SP 9.41

A single phase fully controlled bridge converter supplies a load drawing constant and ripple free load current, if the triggering angle is 30c, the input power factor will be (A) 0.65 (B) 0.78 (C) 0.85 (D) 0.866

EE SP 9.42

A single-phase half controlled converter shown in the figure feeding power to highly inductive load. The converter is operating at a firing angle of 60c.


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Power Electronics

Page 509

If the firing pulses are suddenly removed, the steady state voltage (V0) waveform of the converter will become

EE SP 9.43

nodia

A single phase source inverter is feeding a purely inductive load as shown in the figure. The inverter is operated at 50 Hz in 180c square wave mode. Assume that the load current does not have any dc component. The peak value of the inductor current i0 will be

(A) 6.37 A (C) 20 A

(B) 10 A (D) 40 A

EE SP 9.44

A three phase fully controlled bridge converter is feeding a load drawing a constant and ripple free load current of 10 A at a firing angle of 30c. The approximate Total harmonic Distortion (%THD) and the rms value of fundamental component of input current will respectively be (A) 31% and 6.8 A (B) 31% and 7.8 A (C) 66% and 6.8 A (D) 66% and 7.8 A

EE SP 9.45

A single phase fully controlled converter bridge is used for electrical braking of a separately excited dc motor. The dc motor load is represented by an equivalent circuit as shown in the figure.


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Power Electronics

Chapter 9

Assume that the load inductance is sufficient to ensure continuous and ripple free load current. The firing angle of the bridge for a load current of I0 = 10 A will be (A) 44c (B) 51c (C) 129c (D) 136c EE SP 9.46

nodia

In the circuit shown in the figure, the switch is operated at a duty cycle of 0.5. A large capacitor is connected across the load. The inductor current is assumed to be continuous.

The average voltage across the load and the average current through the diode will respectively be (A) 10 V, 2 A (B) 10 V, 8 A (C) 40 V 2 A (D) 40 V, 8 A YEAR 2007

ONE MARK

EE SP 9.47

A single-phase fully controlled thyristor bridge ac-dc converter is operating at a firing angle of 25c and an overlap angle of 10c with constant dc output current of 20 A. The fundamental power factor (displacement factor) at input ac mains is (A) 0.78 (B) 0.827 (C) 0.866 (D) 0.9

EE SP 9.48

A three-phase, fully controlled thyristor bridge converter is used as line commutated inverter to feed 50 kW power 420 V dc to a three-phase, 415 V(line), 50 Hz ac mains. Consider dc link current to be constant. The rms current of the thyristor is (A) 119.05 A (B) 79.37 A (C) 68.73 A (D) 39.68 A

EE SP 9.49

A single phase full-wave half-controlled bridge converter feeds an inductive load. The two SCRs in the converter are connected to a common DC bus. The converter has to have a freewheeling diode.


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Power Electronics

Page 511

(A) because the converter inherently does not provide for free-wheeling (B) because the converter does not provide for free-wheeling for high values of triggering angles (C) or else the free-wheeling action of the converter will cause shorting of the AC supply (D) or else if a gate pulse to one of the SCRs is missed, it will subsequently cause a high load current in the other SCR. EE SP 9.50

“Six MOSFETs connected in a bridge configuration (having no other power device) must be operated as a Voltage Source Inverter (VSI)”. This statement is (A) True, because being majority carrier devices MOSFETs are voltage driven. (B) True, because MOSFETs hav inherently anti-parallel diodes

nodia

(C) False, because it can be operated both as Current Source Inverter (CSI) or a VSI (D) False, because MOSFETs can be operated as excellent constant current sources in the saturation region.

YEAR 2007

EE SP 9.51

TWO MARKS

A single-phase voltages source inverter is controlled in a single pulse-width modulated mode with a pulse width of 150c in each half cycle. Total harmonic distortion is defined as

2 V rms - V 12 # 100 V1 where V1 is the rms value of the fundamental component of the output voltage. The THD of output ac voltage waveform is (A) 65.65% (B) 48.42% (C) 31.83% (D) 30.49%

THD =

EE SP 9.52

A three-phase, 440 V, 50 Hz ac mains fed thyristor bridge is feeding a 440 V dc, 15 kW, 1500 rpm separately excited dc motor with a ripple free continuos current in the dc link under all operating conditions, Neglecting the losses, the power factor of the ac mains at half the rated speed is (A) 0.354 (B) 0.372 (C) 0.90 (D) 0.955

EE SP 9.53

A single-phase, 230 V, 50 Hz ac mains fed step down transformer (4:1) is supplying power to a half-wave uncontrolled ac-dc converter used for charging a battery (12 V dc) with the series current limiting resistor being 19.04 W. The charging current is (A) 2.43 A (B) 1.65 A (C) 1.22 A (D) 1.0 A

EE SP 9.54

In the circuit of adjacent figure the diode connects the ac source to a pure inductance L.


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Power Electronics

The diode conducts for (A) 90c (C) 270c EE SP 9.55

Chapter 9

(B) 180c (D) 360c

The circuit in the figure is a current commutated dc-dc chopper where, Th M is the main SCR and Th AUX is the auxiliary SCR. The load current is constant at 10 A. Th M is ON. Th AUX is trigged at t = 0 . Th M is turned OFF between.

nodia (A) 0 ms < t # 25 ms (C) 50 ms < t # 75 ms

(B) 25 ms < t # 50 ms (D) 75 ms < t # 100 ms

Common Data for Question 37 and 38. A 1:1 Pulse Transformer (PT) is used to trigger the SCR in the adjacent figure. The SCR is rated at 1.5 kV, 250 A with IL = 250 mA, IH = 150 mA, and IG max = 150 mA, IG min = 100 mA. The SCR is connected to an inductive load, where L = 150 mH in series with a small resistance and the supply voltage is 200 V dc. The forward drops of all transistors/diodes and gate-cathode junction during ON state are 1.0 V

EE SP 9.56

The resistance R should be (A) 4.7 kW (C) 47 W

(B) 470 kW (D) 4.7 W


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Power Electronics

The minimum approximate volt-second rating of pulse transformer suitable for triggering the SCR should be : (volt-second rating is the maximum of product of the voltage and the width of the pulse that may applied) (A) 2000 mV-s (B) 200 mV-s (C) 20 mV-s (D) 2 mV-s YEAR 2006

EE SP 9.58

Page 513

ONE MARK

The speed of a 3-phase, 440 V, 50 Hz induction motor is to be controlled over a wide range from zero speed to 1.5 time the rated speed using a 3-phase voltage source inverter. It is desired to keep the flux in the machine constant in the constant torque region by controlling the terminal voltage as the frequency changes. The inverter output voltage vs frequency characteristic should be

nodia EE SP 9.59

A single-phase half wave uncontrolled converter circuit is shown in figure. A 2-winding transformer is used at the input for isolation. Assuming the load current to be constant and V = Vm sin wt , the current waveform through diode D2 will be


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Power Electronics

Chapter 9


EE SP 9.61

EE SP 9.62

A single-phase inverter is operated in PWM mode generating a single-pulse of width 2d in the centre of each half cycle as shown in figure. It is found that the output voltage is free from 5th harmonic for pulse width 144c. What will be percentage of 3rd harmonic present in the output voltage (Vo3 /Vo1 max) ?

nodia (A) 0.0%

(B) 19.6%

(C) 31.7%

(D) 53.9%

A 3-phase fully controlled bridge converter with free wheeling diode is fed from 400 V, 50 Hz AC source and is operating at a firing angle of 60c. The load current is assumed constant at 10 A due to high load inductance. The input displacement factor (IDF) and the input power factor (IPF) of the converter will be (A) IDF = 0.867; IPF = 0.828 (B) IDF = 0.867; IPF = 0.552 (C) IDF = 0.5; IPF = 0.478 (D) IDF = 0.5; IPF = 0.318 A voltage commutation circuit is shown in figure. If the turn-off time of the SCR is 50 msec and a safety margin of 2 is considered, then what will be the approximate minimum value of capacitor required for proper commutation ?

(A) 2.88 mF (C) 0.91 mF EE SP 9.63

TWO MARKS

(B) 1.44 mF (D) 0.72 mF

A solar cell of 350 V is feeding power to an ac supply of 440 V, 50 Hz through a 3-phase fully controlled bridge converter. A large inductance is connected in the dc circuit to maintain the dc current at 20 A. If the solar cell resistance is 0.5 W ,then each thyristor will be reverse biased for a period of (A) 125c (B) 120c (C) 60c (D) 55c


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EE SP 9.65

Power Electronics

Page 515

A single-phase bridge converter is used to charge a battery of 200 V having an internal resistance of 0.2 W as shown in figure. The SCRs are triggered by a constant dc signal. If SCR 2 gets open circuited, what will be the average charging current ?

(A) 23.8 A

(B) 15 A

(C) 11.9 A

(D) 3.54 A

nodia

An SCR having a turn ON times of 5 msec, latching current of 50 A and holding current of 40 mA is triggered by a short duration pulse and is used in the circuit shown in figure. The minimum pulse width required to turn the SCR ON will be

(A) 251 msec (C) 100 msec

(B) 150 msec (D) 5 msec

Common Data For Q. 66 and 67 A voltage commutated chopper operating at 1 kHz is used to control the speed of dc as shown in figure. The load current is assumed to be constant at 10 A

EE SP 9.66

The minimum time in msec for which the SCR M should be ON is. (A) 280 (B) 140 (C) 70 (D) 0

EE SP 9.67

The average output voltage of the chopper will be (A) 70 V (B) 47.5 V (C) 35 V (D) 0 V


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Power Electronics

Chapter 9

YEAR 2005

ONE MARK

EE SP 9.68

The conduction loss versus device current characteristic of a power MOSFET is best approximated by (A) a parabola (B) a straight line (C) a rectangular hyperbola (D) an exponentially decaying function

EE SP 9.69

A three-phase diode bridge rectifier is fed from a 400 V RMS, 50 Hz, three-phase AC source. If the load is purely resistive, then peak instantaneous output voltage is equal to (A) 400 V (B) 400 2 V (C) 400

EE SP 9.70

2 3

V

(D)

400 3

nodia

The output voltage waveform of a three-phase square-wave inverter contains (A) only even harmonics (B) both odd and even harmonic (C) only odd harmonics (D) only triple harmonics YEAR 2005

EE SP 9.71

TWO MARKS

The figure shows the voltage across a power semiconductor device and the current through the device during a switching transitions. If the transition a turn ON transition or a turn OFF transition ? What is the energy lost during the transition?

(A) Turn ON, VI2 (t1 + t2) (C) Turn ON, VI (t1 + t2) EE SP 9.72

V

(B) Turn OFF, VI (t1 + t2) (D) Turn OFF, VI2 (t1 + t2)

An electronics switch S is required to block voltage of either polarity during its OFF state as shown in the figure (a). This switch is required to conduct in only one direction its ON state as shown in the figure (b)


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Power Electronics

Page 517

Which of the following are valid realizations of the switch S?

(A) Only P

nodia (B) P and Q (C) P and R (D) R and S

EE SP 9.73

The given figure shows a step-down chopper switched at 1 kHz with a duty ratio D = 0.5 . The peak-peak ripple in the load current is close to

(A) 10 A (C) 0.125 A EE SP 9.74

Consider a phase-controlled converter shown in the figure. The thyristor is fired at an angle a in every positive half cycle of the input voltage. If the peak value of the instantaneous output voltage equals 230 V, the firing angle a is close to

(A) 45c (C) 90c EE SP 9.75

(B) 0.5 A (D) 0.25 A

(B) 135c (D) 83.6c

An electric motor, developing a starting torque of 15 Nm, starts with a load torque of 7 Nm on its shaft. If the acceleration at start is 2 rad/sec2 , the moment of inertia of the system must be (neglecting viscous and coulomb friction) (A) 0.25 kg-m2 (B) 0.25 Nm2


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Power Electronics

(C) 4 kg-m2

Chapter 9

(D) 4 Nm2


ONE MARK

A bipolar junction transistor (BJT) is used as a power control switch by biasing it in the cut-off region (OFF state) or in the saturation region (ON state). In the ON state, for the BJT (A) both the base-emitter and base-collector junctions are reverse biased (B) the base-emitter junction is reverse biased, and the base-collector junction is forward biased (C) the base-emitter junction is forward biased, and the base-collector junction is reverse biased

nodia

(D) both the base-emitter and base-collector junctions are forward biased

EE SP 9.77

The circuit in figure shows a full-wave rectifier. The input voltage is 230 V (rms) single-phase ac. The peak reverse voltage across the diodes D 1 and D 2 is

(A) 100 2 V (B) 100 V (C) 50 2 V (D) 50 V EE SP 9.78

The triggering circuit of a thyristor is shown in figure. The thyristor requires a gate current of 10 mA, for guaranteed turn-on. The value of R required for the thyristor to turn on reliably under all conditions of Vb variation is

(A) 10000 W (B) 1600 W (C) 1200 W (D) 800 W


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Power Electronics

Page 519

The circuit in figure shows a 3-phase half-wave rectifier. The source is a symmetrical, 3-phase four-wire system. The line-to-line voltage of the source is 100 V. The supply frequency is 400 Hz. The ripple frequency at the output is

(A) 400 Hz

(B) 800 Hz

(C) 1200 Hz

(D) 2400 Hz

nodia YEAR 2004

EE SP 9.80

A MOSFET rated for 15 A, carries a periodic current as shown in figure. The ON state resistance of the MOSFET is 0.15 W. The average ON state loss in the MOSFET is

(A) 33.8 W (C) 7.5 W EE SP 9.81

(B) 15.0 W (D) 3.8 W

The triac circuit shown in figure controls the ac output power to the resistive load. The peak power dissipation in the load is

(A) 3968 W (C) 7935 W EE SP 9.82

TWO MARKS

(B) 5290 W (D) 10580 W

Figure shows a chopper operating from a 100 V dc input. The duty ratio of the main switch S is 0.8. The load is sufficiently inductive so that the load current is ripple free. The average current through the diode D under steady state is


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EE SP 9.83

Power Electronics

Chapter 9

(A) 1.6 A

(B) 6.4 A

(B) 8.0 A

(D) 10.0 A

Figure shows a chopper. The device S 1 is the main switching device. S 2 is the auxiliary commutation device. S 1 is rated for 400 V, 60 A. S 2 is rated for 400 V, 30 A. The load current is 20 A. The main device operates with a duty ratio of 0.5. The peak current through S 1 is

nodia (A) 10 A (C) 30 A

(B) 20 A (D) 40 A

EE SP 9.84

A single-phase half-controlled rectifier is driving a separately excited dc motor. The dc motor has a back emf constant of 0.5 V/rpm. The armature current is 5 A without any ripple. The armature resistance is 2 W. The converter is working from a 230 V, single-phase ac source with a firing angle of 30c. Under this operating condition, the speed of the motor will be (A) 339 rpm (B) 359 rpm (C) 366 rpm (D) 386 rpm

EE SP 9.85

A variable speed drive rated for 1500 rpm, 40 Nm is reversing under no load. Figure shows the reversing torque and the speed during the transient. The moment of inertia of the drive is


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Power Electronics

(A) 0.048 kg-m2 (C) 0.096 kg-m2

Page 521

(B) 0.064 km-m2 (D) 0.128 kg-m2


ONE MARK

Figure shows a thyristor with the standard terminations of anode (A), cathode (K), gate (G) and the different junctions named J1, J2 and J3. When the thyristor is turned on and conducting

nodia (A) J1 and J2 are forward biased and J3 is reverse biased (B) J1 and J3 are forward biased and J2 is reverse biased (C) J1 is forward biased and J2 and J3 are reverse biased (D) J1, J2 and J3 are all forward biased

EE SP 9.87

Figure shows a MOSFET with an integral body diode. It is employed as a power switching device in the ON and OFF states through appropriate control. The ON and OFF states of the switch are given on the VDS - IS plane by

EE SP 9.88

The speed/torque regimes in a dc motor and the control methods suitable for the


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Power Electronics

Chapter 9

same are given respectively in List-II and List-I List-I

EE SP 9.89

List-II

P.

Field Control

1.

Below base speed

Q.

Armature Control

2.

Above base speed

3.

Above base torque

4.

Below base torque

Codes: (A) P-1, Q-3

(B) P-2, Q-1

(C) P-2, Q-3

(D) P-1, Q-4

A fully controlled natural commutated 3-phase bridge rectifier is operating with a firing angle a = 30c, The peak to peak voltage ripple expressed as a ratio of the peak output dc voltage at the output of the converter bridge is (A) 0.5 (B) 3 /2

nodia (C) c1 - 3 m 2

(D)

3 -1

YEAR 2003

EE SP 9.90

TWO MARKS

A phase-controlled half-controlled single-phase converter is shown in figure. The control angle a = 30c

The output dc voltage wave shape will be as shown in


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EE SP 9.92

Power Electronics

Page 523

A chopper is employed to charge a battery as shown in figure. The charging current is 5 A. The duty ratio is 0.2. The chopper output voltage is also shown in the figure. The peak to peak ripple current in the charging current is

(A) 0.48 A

(B) 1.2 A

(C) 2.4 A

(D) 1 A

nodia

An inverter has a periodic output voltage with the output wave form as shown in figure

When the conduction angle a = 120c, the rms fundamental component of the output voltage is (A) 0.78 V (B) 1.10 V (C) 0.90 V

(D) 1.27 V

EE SP 9.93

With reference to the output wave form given in above figure , the output of the converter will be free from 5 th harmonic when (A) a = 72c (B) a = 36c (C) a = 150c (D) a = 120c

EE SP 9.94

An ac induction motor is used for a speed control application. It is driven from an inverter with a constant V/f control. The motor name-plate details are as follows (no. of poles = 2) V: 415 V VPh: 3 V f: 50 Hz N: 2850 rpm The motor runs with the inverter output frequency set at 40 Hz, and with half the rated slip. The running speed of the motor is (A) 2400 rpm (B) 2280 rpm (C) 2340 rpm (D) 2790 rpm

***********


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Power Electronics

Chapter 9

SOLUTIONS SOL 9.1

Correct answer is 23 ohm. We have the rectifier circuit as

nodia

As the diodes are ideal, so if we are maintaining same current through resistor then Vab also does not change, i.e. Vab = 230 V and Iab = 10 A Hence, the required value of resistance is Rab = Vab = 230 = 23 W 10 Iab

SOL 9.2

Correct option is (C). We have to check all the given switches whether it can block voltages of either polarity when the active device is in the OFF state. Switch (i):

When Va > Vb ; diode will be OFF, but the transistor is ON. When Vb > Va ; transistor is OFF, but diode will be ON. So, the switch can not block voltages of either polarity when the active device (diode or transistor) is in the OFF state. Switch (ii):


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Power Electronics

Page 525

When Va > Vb , both diode and transistor are ON. When Vb > Va both diode and transistor are OFF. So, the switch can block voltages of either polarity when the active devices (diode and transistor) are in the OFF state. Switch (iii):

When Va > Vb , the SCR is ON. When Vb > Va , the SCR is OFF. So, the switch can block voltages of either polarity when the active device (SCR) is in the OFF state. Switch (iv):

nodia

When Va > Vb , diode is OFF. When Vb > Va , SCR is OFF. So, the switch can not block voltages of either polarity when the active device (SCR or diode) is in the OFF state. Thus, the switches (ii) and (iii) only can block voltage of either polarity, when active device is OFF.

SOL 9.3

Correct answer is 69.35°. We have the rectifier circuit as

From the circuit, we obtain p Vo = 1 V sin wtd ^wt h 2p a m = 1 Vm 6- cos wt @Ta 2p = 1 Vm 61 + cos a@ 2p Since, from the given problem, we have Vo = 70 V ; Vm = 325 V Substituting it in equation (i), we get 70 = 1 32.5 61 + cos a@ 2p or 1 + cos a = 70 # 2p 325

#

or or

...(i)

1 + cos a = 1.3526 cos a = 0.3526


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Power Electronics

a = cos-1 ^0.3526h = 69.35c

Thus,

SOL 9.4

Chapter 9

Correct answer is 2.5 A. We have the chopper circuit as

nodia Since, the circuit is a buck regular. So, we have

Vo = Vs # a = 100 # 0.05 0.1

(a = duty cycle)

= 50 V Therefore, peak to peak inductor ripple current is obtained as TI = Vo # Dt L 50 # 0.05 ^m sech = 1 mH = 2.5 A

SOL 9.5

Correct answer is 77.15° Given the output voltage of an inverter as

Since, the output voltage is an odd function, so we compute only sine terms for the given period Vmax

= 2 T

T

# v^t h sin ^w t hdt o

0

From the output wave, we have T = 360c = 2p and w o = 2p = 2p = 1 2p T So, we get the maximum value of fundamental component as 180 - a 180 a Vmax = 2 ; 100 sin tdt - 100 sin tdt + 100 sin tdt 2p 0 a 180 - a

#

#

#


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Power Electronics

#

Page 527 180 + a

- 100 sin tdt + 180

360 - a

360

# 100 sin tdt - # 100 sin tdtE 180 + a

360 - a

= 100 6^1 - cos ah - ^cos a + cos ah + ^1 - cos ah + ^- cos a - cos ah-^- 1 + cos ah@ p = 100 64 - 8 cos a@ p ...(i) = 400 61 - 2 cos a@ p Since, the rms value is Vrms = 50 V Vmax = 50 2 Substituting it in equation (i), we get 400 ^1 - cos ah = 50 p 2 50 p 2 E or cos a = 1 ;1 - # 400 2

nodia a = 77.15c

Thus,

SOL 9.6

Correct answer is 31.2 . As step up chopper

In step-up chopper output voltage is given by where a is duty cycle a = TON Vo = Vs 1-a T 1 - a = Vs = 250 400 Vo 1 - a = 5 & a = 3/8 8 TON = 3 So, 8 T T - TOFF = 3 (T = TON + TOFF ) 8 T 1 - TOFF = 3 8 T TOFF = 5 where, TOFF = 20 ms and T = 1 8 T f Hz So, f # 20 # 10-6 = 5 8 5 = 31.2 kHz f = 8 # 20 # 10-6 SOL 9.7



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Power Electronics

Chapter 9

For a fully controlled bridge converter input power factor is given by Input P.F. = 2 2 cos a p = 2 2 cos 30c p =2 2 # 3 p 2 - 0.779 SOL 9.8

Correct option is (C). The maximum firing angle at which the voltage across device becomes zero is the minimum firing angle to converter. It is given by f = tan-1 b wL l R

nodia -3

= tan-1 b 2p # 50 # 16 # 10 l - 45c 5 Voltage across the device is zero is vo = vi If input and output voltage are identical then Vo rms = 230 V 230 Io rms = Vo rms = 2 2 Z ^5 h + ^2p # 50 # 16 # 10-3h = 230 5 2 rms value of current through SCR 230 Ilrms = Io rms = = 23 A 2 5 2# 2

SOL 9.9


Let us assume SCR is conducting ÎL = 40 mAh 500 R ii ^ t h = Vs 81 - e- L tB = 0.2 81 - e 200 # 10 R

-3

# 50 # 10

B

-6

ii = 0.023 = 23 mA

Rs =

100 = 5882.35 W 17 # 10-3


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SOL 9.10

Power Electronics

Page 529

Correct answer is 0.64 ms. Given the oscillator circuit realiced using a hyteresis type TTL inverter as

nodia Also, we have the upper and lower trigger level voltages as UTL = 1.7 V LTL = 0.9 V For the capacitor, the voltage across it, is given as

Vc (t ) = Vc (3) + [Vc (0) - Vc (3)] e-t/RC For the given data, we have

Vc (0) = 1.7 V Vc (3) = 0 V and Vc (t ) = 0.9 So, we obtain the time t after which output triggers as 0.9 = 0 + [1.7 - 0] e-t/RC 3

0.9 = 1.7e-t/(10 # 10 # 0.1 # 10 t = 0.636 # 10-3 sec

-6

)

= 0.64 ms SOL 9.11

Correct option is (B). When R phase has maximum voltage at conducts and lines a and c conducts. The current are divided into R and YB winding.

R = Io # 2R = 2Io 3 3R I o current through B and Y = 3 Phase an conduct from 90° to 150° current through


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Power Electronics

where B conducts,

Chapter 9

Io = 2I 3

R and Y = Io 3 current through B = 2Io 3 The current through D -phase winding current through

nodia current through primary, I p = kIs

SOL 9.12

Correct answer is 17.07 A. We have the diode circuit as

Given parameters for the circuit are w = 100p rad/s Vs = 100 sin ^wt h V L = 31.83 mH = 31.83 # 10-3 H When switch is closed at t = 2.5 ms , we have Vs - L dI = 0 (in positive half cycle) dt or Vs = L dI dt For the current to be maximum, we have dI = 0 dt or

...(i)

Vs = 0


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Power Electronics

Page 531

Vs = 100 sin wt = 0

So,

wt = np or t = p = 10 ms 100p Therefore, the current is obtained as 1 t V dt = IdI L t s 0 10 ms 10 ms 1 or Vs dt = I =1 100 sin wt + d ^wt h 3 L 2.5 ms 31.83 # 10 2.5 ms 1 cos wt t = 10 ms = -3 # 100 9w Ct = 2.5 ms 31.83 # 10 1 = c 1 + 1m # 100 # p 100 31.83 # 10-3 2 = 17.07 A or

#

#

2

1

#

#

SOL 9.13

nodia Correct option is (D). It is given that load current is sinusoidal, so continuous conduction.

when S1 is off from 0 to q , diode parallel to 53 conducts and VAO =-VDC 2 from q < angle < (p - q), S1 conducts VAO = VOC 2 Again from (p - q) to (q), diode conducts and VAO =-VDC 2

SOL 9.14

Correct option is (C). Given,

L = 100 mH p C = 100 mF p When the circuit is triggered by 10 ms pulse, the thyristor is short circuited and so, we consider IC = Im sin wt Therefore, voltage stored across capacitor is VC = 1 IC dt C

#

= Vm ^1 - cos wt h where w is angular frequency obtained as w = 1 = 100 1 -6 = p # 10 4 ^ p h # 10 LC So, T = 1 = 2p = 200 ms w f As IC = Im sin wt oscillates between - ve and - ve half cycle so, circuit is conducting for only half of cycle and thyristor is open after half cycle. i.e., the conduction period = T = 100 ms 2 SOL 9.15



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Power Electronics

Chapter 9

Given, the rated armature current Ia^rated h = 20 A as rated armature voltage Va^rated h = 150 volt Also, for the armature, we have La = 0.1 mH , Ra = 1 W and T = 50% of Trated So, we get

^T " Torqueh

I = 6Ia^rotatedh@^0.5h = 10 A N = Nrated , I f = I f rated " rated field current

nodia At the rated conditions,

E = V - Ia^ratedh Ra

= 150 - 20 ^1 h = 130 volt

For given torque, Therefore, or,

V = E + Ia Ra = 130 + ^10h^1 h = 140 V chopper output = 140 V D ^200h = 140 D = 140 = 0.7 200

or,

SOL 9.16

(D " duty cycle)

Correct option is (C). Here, as the current from source of 12 V is the same as that pass through inductor. So, the peak to peak current ripple will be equal to peak to peak inductor current. Now, the peak to peak inductor current can be obtained as IL (Peak to Peak) = Vs D Ts L where,

Vs " source voltage = 12 volt , L " inductance = 100mH = 10-4 H , D " Duty ration = 0.4 ,

TS " switching time period of MOSFET = 1 fS and fs " switching frequency = 250 kHz Therefore, we get 1 IL^Peak to Peakh = 12-4 # 0.4 # 250 # 103 10 = 0.192 A This is the peak to peak source current ripple. SOL 9.17

Correct option is (B). Here, the average current through the capacitor will be zero. (since, it is a boost converter). We consider the two cases : Case I : When MOSFET is ON (i 0 is output current) ic =- i 0 1


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Power Electronics

Page 533

(since, diode will be in cut off mode) Case II : When MOSFET is OFF Diode will be forward biased and so ic = Is - i 0 Therefore, average current through capacitor i + Ic Ic, avg = c 2 1

1

0=

&

(Is is source current)

2

DTs ^- io h + ^1 - D h Ts Îs - io h 2

(D is duty ratio)

Solving the equation, we get i0 ^1 - D h Since, the output load current can be given as V/ 12/0.6 = 1A i 0 = V0 = s 1 - D = 20 R R Is =

....(1)

nodia Hence, from Eq. (1)

Is =

SOL 9.18

i0 = 1 = 5 A 0.6 3 1-D

Correct option is (D). We consider the following two cases : Case I : When Q1, Q2 ON

In this case the + ve terminal of V0 will be at higher voltage. i.e. V0 > 0 and so i 0 > 0 (i.e., it will be + ve ). Now, when the Q1 , Q2 goes to OFF condition we consider the second case. Case II : When Q 3 , Q 4 ON and Q , Q2 OFF : In this condition, - ve terminal of applied voltage V0 will be at higher potential i.e., V0 < 0 and since, inductor opposes the change in current so, although the polarity of voltage V0 is inversed, current remains same in inductor i.e. I 0 > 0 . This is the condition when conduction have been asked. In this condition ^V0 > 0, I 0 > 0h since, IGBT’s can’t conduct reverse currents therefore current will flow through D 3, D 4 until ID becomes negative. Thus, D 3 and D 4 conducts.

SOL 9.19

Correct option is (D). When Q 3, Q 4 is switched ON, initially due to the reverse current it remain in OFF state and current passes through diode. In this condition the voltage across Q 3 and Q 4 are zero as diodes conduct. Hence, it shows zero voltage switching during turn-ON

SOL 9.20

Correct option is (D). The circuit of a single-phase half controlled bridge rectifier with RL load and free wheel diode is shown as below.


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Power Electronics

Chapter 9

The voltage current wave forms are shown in figure below.

nodia We note that, for continuous load current, the flywheel diode conducts from p to p + a in a cycle. Thus, fraction of cycle that freewheel diode conducts is a/p. Thus fraction of cycle that freewheel diode conducts is a/p. SOL 9.21

Correct option is (B). The latching current is higher than the holding current. Usually, latching current is taken two to three times the holding currents.

SOL 9.22

Correct option is (C). IS = I 0 + Tic = 5 + 0.8 = 5.8 A 2

SOL 9.23

Correct option is (B). For a three-phase bridge inverter, rms value of output line voltage is


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Power Electronics

VL =

Page 535

2V = 2 300 3 dc 3 #

Vdc = 300 V

= 141.4 V SOL 9.24

Correct option is (D). 2

P = 3 # VL R = 3# SOL 9.25

SOL 9.26

SOL 9.27

(141.4) 2 - 3 kW 20

Correct option is (C). Only option C allow bi direction power flow from source to the drive

nodia

Correct option is (C). Once the SCR start conducting by an forward current, the gate has no control on it and the device can be brought back to the blocking state only by reducing the forward current to a level below that of holding current. This process of turn-off is called commutation. This time is known as the circuit turn-off time of an SCR. Correct option is (A). Maximum current through main thyristor IM (max) = I 0 + Vs C L = 10 + 200

0.1 # 10-6 1 # 103

= 12 A Maximum current through auxiliary thyristor IA (max) = I 0 = 10 A SOL 9.28

Correct option is (A). Output voltage of 3-phase bridge converter V0 = 3 3 Vph cos a p Maximum output (V0) max = 3 3 Vph cos a = 1 p = 3 3 # 400 # 2 = 540.6 V p 3


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Power Electronics

Chapter 9

Resistance of filter choke is 10 W, So (V0) max = E + IR chock 540.6 = 400 + I (10) I - 14 A SOL 9.29

Correct option is (D). kVA rating =

3 VL IL =

3 # 400 # 6 # 14 p

= 7.5 kVA SOL 9.30

Correct option is (A). The figure shows a step down chopper circuit.

nodia Vout = DVin where, D = Duty cycle and D < 1

SOL 9.31

Correct option is (C). Given figure as

The I -V characteristic are as

Since diode connected in series so I can never be negative. When current flows voltage across switch is zero and when current is zero than there may be any voltage across switch. SOL 9.32

Correct option is (A). Given fully-controlled thyristor converter, when firing angle a = 0 , dc output voltage Vdc = 300 V If a = 60c, then Vdc = ? For fully-controlled converter 0

Vdc = 0

2 2 Vdc cos a p 1

Since a = 0 , Vdc = 300 V 0

2 2 Vdc cos 0c p = 300p 2 2

300 = Vdc At a

1

1

= 60c, Vdc = ? 2

Vdc = 2 2 # 300p cos 60c p 2 2 = 300 # 1 = 150 V 2 2


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Power Electronics

Page 537

SOL 9.33

Correct option is (C). SCR has the property that it can be turned ON but not OFF with a gate pulse, So SCR is being considered to be a semi-controlled device.

SOL 9.34


Current wave form for iL vL = LdiL dt iL = 1 # vL dt 2

nodia for 0 < wt +p,

vL = vin = 10 sin wt = diL dt iL = 1 # vL dt =- cos 100pt + C 2

at 100pt = p/2 ,

iL = 0 , C = 0 iL =- 100 cos pt

iL (peak) = 1 Amp

SOL 9.35

for p < wt vL = vin = 0

Correct option is (C). In CSI let T3 and T4 already conducting at t = 0 At triggering T1 and T2 , T3 and T4 are force cumulated. Again at t = T , T1 and T2 are force cumulated. This completes a cycle. 2

Time constant t = RC = 4 # 0.5 = 2 m sec 1 Frequency f = 1 = = 500 kHz t 2 # 10- 6


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Power Electronics

Chapter 9

Correct option is (A). duty ratio TM = 0.8 Maximum dv on TM = 50 V/msec dt Minimum value of C1 = ? Given that current ripple through L 0 is negligible. Current through TM = Im = duty ratio # current = 0.8 # 12.5 = 10 A Im = C1 dv dt 10 = C1 # 50- 6 10 C1 = 50 # 10- 6 = 0.2 mF 10

a

SOL 9.37

SOL 9.38

nodia Correct option is (C). Characteristics are as



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Power Electronics

Page 539

R + jXL = 50 + 50j tan f = wL = 50 = 1 50 R

`

f = 45c so, firing angle ‘a’ must be higher the 45c, Thus for 0 < a < 45c, V0 is uncontrollable. SOL 9.39

Correct option is (D). A 3-f voltage source inverter is operated in 180c mode in that case third harmonics are absent in pole voltage and line voltage due to the factor cos (np/6) . so both are free from 3rd harmonic components.

SOL 9.40


nodia 1 TON1 + TON 2 TON 2 D = TON1 + TON 2 f =

In this case and,

SOL 9.41

Correct option is (B). Given a = 30c, in a 1-f fully bridge converter we know that, Power factor = Distortion factor # cos a D.f. (Distortion factor) = Is(fundamental) /Is = 0.9 power factor = 0.9 # cos 30c = 0.78

SOL 9.42

Correct option is (A). Output of this

Here the inductor makes T1 and T3 in ON because current passing through T1 and T3 is more than the holding current. SOL 9.43

Correct option is (C). Input is given as


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Power Electronics

Chapter 9

Here load current does not have any dc component ` `

Peak current occur at (p/w) Vs = L di dt 200 = 0.1 # di dt di = a p kb 1 l = 1 2p 50 100 di(max) = 200 # 1 # 1 100 0.1

nodia Here So

= 20 A

SOL 9.44


Total rms current Ia =

2 10 = 8.16 A 3#

Fundamental current Ia1 = 0.78 # 10 = 7.8 A THD =

where

1 -1 DF2

DF = Ia1 = 0.78 # 10 = 0.955 0.816 # 10 Ia ` SOL 9.45

THD =

1 2 b 0.955 l - 1 = 31%

Correct option is (C). Here for continuous conduction mode, by Kirchoff’s voltage law, average load current

V - 2Ia + 150 = 0 Ia = V + 150 2 ` I1 = 10 A, So V =- 130 V 2Vm cos a =- 130 p


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Power Electronics

2#

Page 541

2 # 230 cos a =- 130c p a = 129c

SOL 9.46


In the given diagram when switch S is open I 0 = IL = 4 A, Vs = 20 V when switch S is closed ID = 0, V0 = 0 V Duty cycle = 0.5 so average voltage is Vs 1-d Average current = 0 + 4 = 2 amp 2 Average voltage = 20 = 40 V 1 - 0.5

SOL 9.47

nodia Correct option is (A). Firing angle Overlap angle so,

20 =

` `

a = 25c m = 10c I 0 = Vm [cos a - cos (a + m)] wLs

230 2 [cos 25c - cos (25c + 10c)] 2p # 50Ls

Ls = 0.0045 H V0 = 2Vm cos a - wLsI 0 p p -3 = 2 # 230 2 cos 25c - 2 # 3.14 # 50 # 4.5 # 10 # 20 3.14 3.14

= 187.73 - 9 = 178.74c Displacement factor = V0 I 0 = 178.25 # 20 = 0.78 230 # 20 Vs Is SOL 9.48

Correct option is (C). Given that P = 50 # 1000 W Vd = 420 So P = Vd # Id Id = 50 # 1000 = 119.05 420 RMS value of thyristor current = 119.05 = 68.73 3


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Power Electronics

Chapter 9

SOL 9.49

Correct option is (B). Single phase full wave half controlled bridge converter feeds an Inductive load. The two SCRs in the converter are connected to a common dc bus. The converter has to have free wheeling diode because the converter does not provide for free wheeling for high values of triggering angles.

SOL 9.50

Correct option is (D). If we connect the MOSFET with the VSI, but the six MOSFETs are connected in bridge configuration, in that case they also operated as constant current sources in the saturation region so this statement is false.

SOL 9.51

Correct option is (C). Given that, total harmonic distortion

nodia THD =

Vrms - V 12 # 100 V1 2

Pulse width is 150c

Vrms = b

Here

150 V = 0.91V s 180 l s

V1 = Vrms(fundamental) = 0.4Vs sin 75c p# 2 = 0.8696Vs

(0.91Vs) 2 - (0.87Vs) 2 (0.87Vs) 2 = 31.9%

THD =

SOL 9.52

Correct option is (A). When losses are neglected, 2 # 440 cos a = K 750 # 2p m# p 60 Here back emf e with f is constant 3#

e = V0 = Km wm 440 = Km # 1500 # 2p 60 Km = 2.8 cos a = 0.37 at this firing angle Vt = 3 2 # 440 # (0.37) p = 219.85 V Ia = 1500 = 34.090 440 Isr = Ia 2/3 = 27.83 p.f. =

Vt Is = 0.354 3 Vs Isr


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Power Electronics

Page 543

Correct option is (D). Vs = 230 = 57.5 4 Here charging current = I Vm sin q = 12 q1 = 8.486 = 0.148 radian Vm = 81.317 V e = 12 V There is no power consumption in battery due to ac current, so average value of charging current. 1 [2Vm cos q1 - e (p - 2q1)] Iav(charging) = 2p # 19.04 1 [2 V cos q1 - 12 (p - 2q1)] = 2p # 19.04 # m #

nodia = 1.059 W/A

SOL 9.54

Correct option is (C). Conduction angle for diode is 270c as shown in fig.

SOL 9.55


SOL 9.56

Correct option is (C). Here, Vm = maximum pulse voltage that can be applied so = 10 - 1 - 1 - 1 = 7 V Here 1 V drop is in primary transistor side, so that we get 9V pulse on the secondary side. Again there are 1 V drop in diode and in gate cathode junction each. So

SOL 9.57

I g max = 150 mA 7 = 46.67 W R = Vm = Ig max 150 mA

Correct option is (A). We know that the pulse width required is equal to the time taken by ia to rise upto iL so, Vs = L di + Ri (VT . 0) dt ia = 200 [1 - e- t/0.15] 1 Here also

t = T,

ia = iL = 0.25


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Power Electronics

Chapter 9

0.25 = 200 [1 - e- T/0.5] T = 1.876 # 10- 4 = 187.6 ms Width of pulse = 187.6 ms Magnitude of voltage = 10 V Vsec rating of P.T. = 10 # 187.6 ms = 1867 mV-s is approx to 2000 mV-s SOL 9.58

Correct option is (D). If we varying the frequency for speed control, V/f should be kept as constant so that, minimum flux density (Bm ) also remains constant V = 4.44NBm Af

So,

SOL 9.59

SOL 9.60

nodia

Correct option is (D). In first half cycle D 1 will conduct and D 2 will not and at q = 0 there is zero voltage. So current wave form is as following

Correct option is (B). In the PWM inverter V0 = output voltage of inverter 3 V0 = / 4Vs sin nd sin nwt sin np/2 n = 1 np

so, SOL 9.61

So the pulse width = 2d = 144c V01 = 4Vs sin 72c sin wt p V03 = 4Vs sin ^3 # 72ch sin 3wt 3p 4Vs sin (3 # 72c) V03 3p # = = 19.61% bV01 max l 4Vs sin 72c p

Correct option is (C). Given that 400 V, 50 Hz AC source, so,

a = 60c, IL = 10 A

Input displacement factor = cos a = 0.5


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SOL 9.62

Power Electronics

and,

input power factor = D.F. # cos a

so,

4 # 10 sin 60c Is(fundamental) p distortion factor = = # 2 Is 10 # 2/3 = 0.955 input power factor = 0.955 # 0.5 = 0.478

Correct option is (A). We know that So

SOL 9.63

Page 545

T = RC ln 2 T 100 = = 2.88 mF C = 50 # 0.693 R # 0.693

nodia Correct option is (A). Let we have so

`

R solar = 0.5 W , I 0 = 20 A Vs = 350 - 20 # 0.5 = 340 V 340 = 3 # 440 # p

2 cos a

cos a = 55c So each thyristor will reverse biased for 180c - 55c = 125c.

SOL 9.64

Correct option is (C). In this circuitry if SCR gets open circuited, than circuit behaves like a half wave rectifier.

So

I avg = Average value of current p-q = 1 # (Vm sin wt - E) dq 2pR q I 0(avg) = 1 62Vm cos q - E (p - 2q1)@ 2pR 1 [2 (230 = # 2 ) cos q - 200 (p - 2q1)] 2p # 2 # 200 q1 = sin- 1 b E l = sin- 1 c = 38c = 0.66 Rad Vm 230 # 2 m 1

1

a


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Power Electronics

`

I 0 (avg) =

Chapter 9

1 [2 2 # 230 cos 38c - 200 (p - 2 # 0.66)] 2p # 2

= 11.9 A SOL 9.65


In this given circuit minimum gate pulse width time= Time required by ia rise up to iL i2 = 100 3 = 20 mA 5 # 10 i1 = 100 [1 - e- 40t] 20

nodia `

anode current I = I1 + I2 = 0.02 + 5 [1 - e- 40t] 0.05 = 0.05 + 5 [1 - e- 40t] 1 - e- 40t = 0.03 5 T = 150 ms

SOL 9.66

Correct option is (B). Given IL = 10 A . So in the + ve half cycle, it will charge the capacitor, minimum time will be half the time for one cycle. so min time required for charging = p = p LC = 3.14 # w0

SOL 9.67

2 # 10- 3 # 10- 6 = 140 m sec

Correct option is (C). Given

Ton = 140 m sec Average output = Ton # V Ttotal Ttotal = 1/f = 1 = 1 msec 103 -6

so average output = 140 # 10-3 # 250 = 35 V 1 # 10 SOL 9.68

Correct option is (A). The conduction loss v/s MOSFET current characteristics of a power MOSFET is best approximated by a parabola.

SOL 9.69

Correct option is (B). In a 3-f bridge rectifier Vrms = 400 V , f = 50 Hz This is purely resistive then instantaneous voltage

V0 =

2 Vrms = 400 2 V


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Power Electronics

Page 547

SOL 9.70

Correct option is (C). A 3-f square wave (symmetrical) inverter contains only odd harmonics.

SOL 9.71

Correct option is (A). In Ideal condition we take voltage across the device is zero. average power loss during switching = VI (t1 + t2) (turn ON) 2

SOL 9.72

Correct option is (C). So in P thyristor blocks voltage in both polarities until gate is triggered and also in R transistor along with diode can do same process.

SOL 9.73


nodia Duty ratio a = 0.5

1 = 10- 3 sec 1 # 10- 3 Ta = L = 200 mH = 40 msec 5 R T =

here

(1 - e- aT/Ts) (1 - e- (1 - a) T/Ta) Ripple = Vs = G R 1 - e- T/Ts 100 (TI) max = Vs = 4fL 4 # 103 # 200 # 10- 3 = 0.125 A

SOL 9.74

Correct option is (B). We know that Vrms = 230 V Vm = 230 # 2 V a 1 90c Vpeak = Vm sin a = 230

so, If whether Then

230 2 sin a = 230 sin a = 1 2 angle a = 135c SOL 9.75

Correct option is (C). Tst = 15 Nm

so

SOL 9.76

= 7 Nm = 2 rad/sec2 = Ia = Tst - TL = 8 Nm I = 8 = 4 kgm2 2

TL a T T

Correct option is (D). When we use BJT as a power control switch by biasing it in cut-off region or in the saturation region. In the on state both the base emitter and base-collector junction are forward biased.


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Power Electronics

Chapter 9

Correct option is (A). Peak Inverse Voltage (PIV) across full wave rectifier is 2Vm Vm = 50 2 V PIV = 100 2 V

so, SOL 9.78

Correct option is (D). Vb = 12 ! 4 V Vb max = 16 V Vb min = 8 V V (min) 8 Required value of R = b = = 800 W Ig 10 # 10- 3

SOL 9.79


Ripple frequency = 3f = 3 # 400 = 1200 Hz So from V0 ripple frequency = 1200 Hz

SOL 9.80

Correct option is (C). Given that So average power losses

SOL 9.81

R = 0.15 W I = 15 A 1 # p/w I 2 Rdt = (2p/w) 0 = w # 102 # 0.15 # p/w = 7.5 W 2p

Correct option is (D). Output dc voltage across load is given as following 1

Vdc =

2 V ; 1 &(2p - a) + sin 2a 0E2 ap 2 1

sin p/2 2 = 2 # 230 2 = p 1 'a2p - p k + b = 317.8 V 2 l1G 4 4 #p 2 (317.8) 2 losses = V dc = = 10100 W 100 R SOL 9.82

Correct option is (C). Vs = 100 V , duty ratio = 0.8 , R = 10 W So average current through diode

SOL 9.83

= aVs = 0.8 # 100 = 8 A 10 R

Correct option is (D). Peak current through S 1


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Power Electronics

I = I 0 + VS C/L = 20 + 200 SOL 9.84


SOL 9.85


Page 549

2 # 10- 6 = 40 A 200 # 10- 6

nodia so

a =;

and

500 - (- 1500) 2p 2 E # 60 = 418.67 rad/sec 0.5

T = 40 Nm

T = Ia I = T # 40 = 0.096 kgm2 a 418.67

SOL 9.86

SOL 9.87

Correct option is (D). When thyristor turned on at that time J2 junction will break. So J1, J2, J3 all are in forward bias. Correct option is (D). The ON-OFF state of switch is given on VDS - IS plane as following

When VDS =+ ve , diode conducts and IS = 0 VDS =- ve , diode opens, but IS = 0 , D "- ve potential. SOL 9.88

Correct option is (B). P. Field control-Above base speed Q. Armature control-below base torque

SOL 9.89

Correct option is (A). As we know in fully controlled rectifier. or or

VPP = Vm - Vm cos (p/6 + a) VPP = Vm [1 - cos (p/6 + 30c)] VPP = 0.5 Vm

a a = 30c


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Power Electronics

SOL 9.90


SOL 9.91

Correct option is (A). In the chopper during turn on of chopper V -t area across L is, T T di dt = # imax Ldi #0 onVL dt = #0 on L b dt l i min

Chapter 9

= L (i max - i min) = L ^DI h

nodia V -t are applied to ‘L’ is = (60 - 12) Ton = 48Ton DI = 48Ton L

So now volt area

-3 = 48 # 0.2 #-10 = 0.48 A 3 20 # 10

SOL 9.92


4VS b np l^sin nd h^sin nwt h^sin np/2h ` RMS value of fundamental component Vrms(fundamental) = 4VS sin d # 1 2p a = 120c, 2d = 120c & d = 60c Output voltage V0 =

3

/

n = 1, 3, 5

Vrms(fundamental) = 4VS # sin 60c 2p = 0.78VS = 0.78 V SOL 9.93



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Power Electronics

Page 551

After removing 5 th harmonic 5d = 0, p, 2p ` SOL 9.94

Pulse width = 2d = a = 0, 2p , 4p = 0c, 72c, 144c 5 5

Correct option is (C). NSa = 3000 rpm Na = 2850 rpm SFL = 3000 - 2850 = 0.05 3000 where by (V/f) control Nsb = 3000 b 40 l = 2400 rpm 50

nodia

`

N2 = new running speed of motor = 2400 b1 - 0.05 l = 2340 rpm 2 ***********


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Chapter 10

CHAPTER 10 ENGINEERING MATHEMATICS


ONE MARK

Given a system of equations x + 2y + 2z = b1 5x + y + 3z = b2 What of the following is true regarding its solutions (A) The system has a unique solution for any given b1 and b2 (B) The system will have infinitely many solutions for any given b1 and b2 (C) Whether or not a solution exists depends on the given b1 and b2

nodia (D) The systems would have no solution for any values of b1 and b2

EE SP 10.2

EE SP 10.3

Let f ^x h = xe-x . The maximum value of the function in the interval ^0, 3h is (A) e-1 (B) e -1 (C) 1 - e (D) 1 + e-1 The solution for the differential equation d 2 x =- 9x dt2 with initial conditions x ^0 h = 1 and dx = 1, is dt t = 0 (B) sin 3t + 13 cos 3t + 23 (A) t2 + t + 1 (C)

EE SP 10.4

EE SP 10.5

1 3

sin 3t + cos 3t

(D) cos 3t + t

Let S be the set of points in the complex plane corresponding to the unit circle. (That is, S = "z : z = 1, ). Consider the function f ^z h = zz * where z* denotes the complex conjugate of z . The f ^z h maps S to which one of the following in the complex plane (A) unit circle (B) horizontal axis line segment from origin to (1, 0) (C) the point (1, 0) (D) the entire horizontal axis x ^ t h is nonzero only for Tx < t < Txl, and similarly, y ^ t h is nonzero only for Ty < t < Tyl. Let z ^ t h be convolution of x ^ t h and y ^ t h. Which one of the following statements is TRUE ? (A) z ^ t h can be nonzero over an unbounded interval. (B) z ^ t h is nonzero for t < Tx + Ty (C) z ^ t h is zero outside of Tx + Ty < t < Txl+ Tyl (D) z ^ t h is nonzero for t > Txl+ Tyl


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Page 553

TWO MARKS

Let g : 60, 3h " 60, 3h be a function defined by g ^x h = x - 6x @, where 6x @ represents the integer part of x . (That is, it is the largest integer which is less than or equal to x ). The value of the constant term in the Fourier series expansion of g ^x h is____.

EE SP 10.7

A fair coin is tossed n times. The probability that the difference between the number of heads and tails is ^n - 3h is (A) 2-n (B) 0 n -n (C) Cn - 3 2 (D) 2-n + 3

EE SP 10.8

The line integral of function F = yzi , in the counterclockwise direction, along the circle x2 + y2 = 1 at z = 1 is (A) - 2p (B) - p (C) p (D) 2p

EE SP 10.9

EE SP 10.10

nodia 1 be the z -transform of a causal signal x 6n@. 1 - z-3 Then, the values of x 62@ and x 63@ are (A) 0 and 0 (B) 0 and 1 (C) 1 and 0 (D) 1 and 1 Let X ^z h =

A system matrix is given as follows V R 1 - 1W S 0 A = S- 6 - 11 6W SS- 6 - 11 5WW X T The absolute value of the ratio of the maximum eigenvalue to the minimum eigenvalue is _______. YEAR 2014 EE02

ONE MARK

EE SP 10.11

Which one of the following statements is true for all real symmetric matrices ? (A) All the eigenvalues are real (B) All the eigenvalues are positive (C) All the eigenvalues are distinct (D) Sum of all the eigenvalues is zero

EE SP 10.12

Consider a dice with the property that the probability of a free with n dots showing up proportional to n . The probability of the face with three dots showing up is____.

EE SP 10.13

Minimum of the real valued function f ^x h = ^x - 1h2/3 occurs at x equal to (A) - 3 (B) 0 (C) 1 (D) 3


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EE SP 10.15


Chapter 10

All the values of the multi-valued complex function 1i , where i = - 1 , are (A) purely imaginary (B) real and non-negative (C) on the unit circle (D) equal in real and imaginary parts d 2y dy Consider the differential equation x2 2 + x - y = 0 . Which of the following dx dx is a solution to this differential equation for x > 0 ? (B) x2 (A) ex (C) 1/x (D) ln x YEAR 2014 EE02

EE SP 10.16

TWO MARKS

nodia To evaluate the double integral substitution

8

# e# 0

^y/2h + 1

y/2

b

2x - y dx dy , we make the 2 l o

u = ^ 2x 2- y h and v = y2 . The integral will reduce to

(A)

(C)

EE SP 10.17

EE SP 10.18

4

2

# c # 2udu mdv 0

0

4

1

# c # udu mdv 0

0

(B)

(D)

4

1

# c # 2udu mdv 0

0

4

2

# c # udu mdv 0

0

Let X be a random variable with probability density function Z0.2, for x # 1 ] f ^x h = [0.1, for 1 < x # 4 ] 0, otherwise The probability P ^0.5 < X < 5\h is _____.

The minimum value of the function f ^x h = x3 - 3x2 - 24x + 100 in the interval 6- 3, 3@ is (A) 20 (B) 28 (C) 16 (D) 32 YEAR 2014 EE03

ONE MARK

EE SP 10.19

Two matrices A and B are given below : p q p2 + q2 pr + qs ; B=> A => H H r s pr + qs r2 + s2 If the rank of matrix A is N , then the rank of matrix B is (A) N/2 (B) N - 1 (C) N (D) 2N

EE SP 10.20

A particle, starting from origin at t = 0 s , is traveling along x -axis with velocity v = p cos a p t k m/s 2 2 At t = 3 s , the difference between the distance covered by the particle and the magnitude of displacement from the origin is ______.


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Page 555

Lifetime of an electric bulb is a random variable with density f ^x h = kx2 , where x is measured in years. If the minimum and maximum lifetimes of bulb are 1 and 2 years respectively, then the value of k is _____. YEAR 2014 EE03

EE SP 10.22

EE SP 10.23

TWO MARKS

2 Integration of the complex function f ^z h = 2 z , in the counterclockwise z -1 direction, around z - 1 = 1, is (A) - pi (B) 0 (C) pi (D) 2pi

The mean thickness and variance of silicon steel laminations are 0.2 mm and 0.02 respectively. The varnish insulation is applied on both the sides of the laminations. The mean thickness of one side insulation and its variance are 0.1 mm and 0.01 respectively. If the transformer core is made using 100 such varnish coated laminations, the mean thickness and variance of the core respectively are (A) 30 mm and 0.22 (B) 30 mm and 2.44

nodia (C) 40 mm and 2.44

(D) 40 mm and 0.24

EE SP 10.24

The function f ^x h = ex - 1 is to be solved using Newton-Raphson method. If the initial value of x 0 is taken as 1.0, then the absolute error observed at 2 nd iteration is _____. YEAR 2013

EE SP 10.25

EE SP 10.26

ONE MARK

Given a vector field Fv = y2 xavx - yzavy - x2 avz , the line integral along a segment on the x -axis from x = 1 to x = 2 is (B) 0 (A) - 2.33 (D) 7 (C) 2.33 2 - 2 x1 0 The equation > = > H has H > H 1 - 1 x2 0 (A) no solution (C) non-zero unique solution

EE SP 10.27

# Fv : dlv evaluated

x1 0 (B) only one solution > H = > H x2 0 (D) multiple solutions

Square roots of - i , where i = - 1 , are (A) i , - i (B) cos d- p n + i sin d- p n, cos b 3p l + i sin b 3p l 4 4 4 4 (C) cos d p n + i sin b 3p l, cos b 3p l + i sin d p n 4 4 4 4 (D) cos b 3p l + i sin b- 3p l, cos b- 3p l + i sin b 3p l 4 4 4 4


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Chapter 10

The curl of the gradient of the scalar field defined by V = 2x2 y + 3y2 z + 4z2 x is (A) 4xyavx + 6yzavy + 8zxavz (B) 4avx + 6avy + 8avz (C) ^4xy + 4z2h avx + ^2x2 + 6yz h avy + ^3y2 + 8zx h avz (D) 0

EE SP 10.29

A continuous random variable X has a probability density function f ^x h = e-x , 0 < x < 3. Then P "X > 1, is (A) 0.368 (B) 0.5 (C) 0.632 (D) 1.0 YEAR 2013

EE SP 10.30

EE SP 10.31

nodia

When the Newton-Raphson method is applied to solve the equation f ^x h = x3 + 2x - 1 = 0 , the solution at the end of the first iteration with the initial value as x 0 = 1.2 is (A) - 0.82 (B) 0.49 (D) 1.69 (C) 0.705 A function y = 5x2 + 10x is defined over an open interval x = ^1, 2h. Atleast at one point in this interval, dy/dx is exactly (A) 20 (B) 25 (C) 30

EE SP 10.32

# zz

, is

2 2

(D) 35

- 4 dz evaluated anticlockwise around the circle z - i = 2 , where i = +4

(A) - 4p (C) 2 + p

EE SP 10.33

TWO MARKS

(B) 0 (D) 2 + 2i

1 A Matrix has eigenvalues - 1 and - 2 . The corresponding eigenvectors are > H -1 1 and > H respectively. The matrix is -2 1 1 (A) > - 1 - 2H

1 2 (B) > - 2 - 4H

-1 0 (C) > 0 - 2H

0 1 (D) > - 2 - 3H

YEAR 2012 EE SP 10.34

-1

ONE MARK

Two independent random variables X and Y are uniformly distributed in the interval 6- 1, 1@. The probability that max 6X, Y @ is less than 1/2 is (A) 3/4 (B) 9/16 (C) 1/4 (D) 2/3


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EE SP 10.36


If x = - 1, then the value of xx is (A) e- p/2 (C) x

Page 557

(B) e p/2 (D) 1

1 - 2 . If C is a counter clockwise path in the z -plane z+1 z+3 such that z + 1 = 1, the value of 1 f (z) dz is 2 pj C (A) - 2 (B) - 1 (C) 1 (D) 2

Given f (z) =

#

EE SP 10.37

With initial condition x (1) = 0.5 , the solution of the differential equation t dx + x = t , is dt (A) x = t - 1 (B) x = t 2 - 1 2 2

nodia 2

(C) x = t 2

(D) x = t 2

YEAR 2012

EE SP 10.38

EE SP 10.39

TWO MARKS

-5 -3 1 0 Given that A = > , the value of A3 is and I = > H 2 0 0 1H (A) 15A + 12I (B) 19A + 30I (C) 17A + 15I (D) 17A + 21I

The maximum value of f (x) = x3 - 9x2 + 24x + 5 in the interval [1, 6] is (A) 21 (B) 25 (C) 41

(D) 46

EE SP 10.40

A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is (A) 1/3 (B) 1/2 (C) 2/3 (D) 3/4

EE SP 10.41

The direction of vector A is radially outward from the origin, with A = krn . where r2 = x2 + y2 + z2 and k is a constant. The value of n for which d:A = 0 is (A) - 2 (B) 2 (C) 1 (D) 0

EE SP 10.42

Consider the differential equation d 2 y (t) dy (t) dy +2 + y (t) = d (t) with y (t) t = 0 =- 2 and 2 dt dt dt dy The numerical value of is dt t = 0 (A) - 2 (B) - 1 (C) 0 (D) 1 -

=0 t = 0-

+


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Chapter 10

YEAR 2011 EE SP 10.43

Roots of the algebraic equation x3 + x2 + x + 1 = 0 are (A) (+ 1, + j, - j) (C) (0, 0, 0)

EE SP 10.44

ONE MARK

(B) (+ 1, - 1, + 1) (D) (- 1, + j, - j)

A point Z has been plotted in the complex plane, as shown in figure below.

nodia The plot of the complex number Y = 1 is Z

EE SP 10.45

With K as a constant, the possible solution for the first order differential equation dy = e-3x is dx (B) - 13 e3x + K (A) - 13 e-3x + K (C) - 13 e-3x + K (D) - 3e-x + K YEAR 2011

EE SP 10.46

TWO MARKS

Solution of the variables x1 and x2 for the following equations is to be obtained by employing the Newton-Raphson iterative method. Equation (1) 10x2 sin x1 - 0.8 = 0 Equation (2) 10x 22 - 10x2 cos x1 - 0.6 = 0


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Page 559

Assuming the initial values are x1 = 0.0 and x2 = 1.0 , the jacobian matrix is 10 (A) > 0 0 (C) > 10 EE SP 10.47

- 0.8 - 0.6H - 0.8 - 0.6H

10 0 (B) > 0 10H 10 0 (D) > 10 - 10H

The function f (x) = 2x - x2 - x3 + 3 has (A) a maxima at x = 1 and minimum at x = 5 (B) a maxima at x = 1 and minimum at x =- 5 (C) only maxima at x = 1 and (D) only a minimum at x = 5

EE SP 10.48

EE SP 10.49

EE SP 10.50

nodia

A zero mean random signal is uniformly distributed between limits - a and + a and its mean square value is equal to its variance. Then the r.m.s value of the signal is (A) a (B) a 3 2 (C) a 2 (D) a 3

2 1 The matrix [A] = > is decomposed into a product of a lower triangular 4 - 1H matrix [L] and an upper triangular matrix [U]. The properly decomposed [L] and [U] matrices respectively are 1 1 2 0 1 1 1 0 and > (B) > and > H (A) > H H H 0 -2 4 -1 0 1 4 -1 1 0 2 1 2 0 1 1.5 (C) > H and > (D) > and > H H 4 1 0 -1 4 -3 0 1H The two vectors [1,1,1] and [1, a, a2] where a = _- 12 + j

3 2

i, are

(A) Orthonormal (B) Orthogonal (C) Parallel (D) Collinear YEAR 2010

ONE MARK 1

EE SP 10.51

The value of the quantity P , where P = # xex dx , is equal to 0 (A) 0 (B) 1 (C) e (D) 1/e

EE SP 10.52

Divergence of the three-dimensional radial vector field r is (A) 3 (B) 1/r t (C) ti + tj + k


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Chapter 10

t) (D) 3 (ti + tj + k YEAR 2010 EE SP 10.53

EE SP 10.54

EE SP 10.55

EE SP 10.56

EE SP 10.57

TWO MARKS

A box contains 4 white balls and 3 red balls. In succession, two balls are randomly and removed form the box. Given that the first removed ball is white, the probability that the second removed ball is red is (A) 1/3 (B) 3/7 (C) 1/2

(D) 4/7

J1 1 0N O K An eigenvector of P = K0 2 2O is K0 0 3O (A) 8- 1 1 1BT P L T (C) 81 - 1 2B

(B) 81 2 1BT (D) 82 1 - 1BT

nodia

2 For the differential equation d x2 + 6 dx + 8x = 0 with initial conditions x (0) = 1 dt dt and dx = 0 , the solution is dt t = 0 (B) x (t) = 2e- 2t - e- 4t (A) x (t) = 2e- 6t - e- 2t (C) x (t) =- e- 6t + 2e- 4t (D) x (t) = e- 2t + 2e- 4t

For the set of equations, x1 + 2x2 + x 3 + 4x 4 = 2 and 3x1 + 6x2 + 3x 3 + 12x 4 = 6 . The following statement is true. (A) Only the trivial solution x1 = x2 = x 3 = x 4 = 0 exists (B) There are no solutions (C) A unique non-trivial solution exists (D) Multiple non-trivial solutions exist

At t = 0 , the function f (t) = sin t has t (A) a minimum (C) a point of inflection

(B) a discontinuity (D) a maximum

YEAR 2009 EE SP 10.58

ONE MARK

The trace and determinant of a 2 # 2 matrix are known to be - 2 and - 35 respectively. Its eigen values are (A) - 30 and - 5 (B) - 37 and - 1 (C) - 7 and 5 (D) 17.5 and - 2 YEAR 2009

EE SP 10.59

TWO MARKS

f (x, y) is a continuous function defined over (x, y) ! [0, 1] # [0, 1]. Given the two constraints, x > y2 and y > x2 , the volume under f (x, y) is (A)

y=1

x= y

#y = 0 #x = y

2

f (x, y) dxdy

(B)

y=1

x=1

#y = x #x = y 2

2

f (x, y) dxdy


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(C)

y=1

x=1

#y = 0 #x = 0

f (x, y) dxdy

(D)

Page 561

y= x

#y = 0

x= y

#x = 0

f (x, y) dxdy

EE SP 10.60

Assume for simplicity that N people, all born in April (a month of 30 days), are collected in a room. Consider the event of at least two people in the room being born on the same date of the month, even if in different years, e.g. 1980 and 1985. What is the smallest N so that the probability of this event exceeds 0.5 ? (B) 7 (A) 20 (C) 15 (D) 16

EE SP 10.61

A cubic polynomial with real coefficients (A) Can possibly have no extrema and no zero crossings

nodia

(B) May have up to three extrema and upto 2 zero crossings (C) Cannot have more than two extrema and more than three zero crossings (D) Will always have an equal number of extrema and zero crossings

EE SP 10.62

EE SP 10.63

Let x2 - 117 = 0 . The iterative steps for the solution using Newton-Raphon’s method is given by (A) xk + 1 = 1 bxk + 117 l (B) xk + 1 = xk - 117 2 xk xk (C) xk + 1 = xk - xk (D) xk + 1 = xk - 1 bxk + 117 l 2 xk 117

F (x, y) = (x2 + xy) at x + (y2 + xy) at y . It’s (x, y) = (0, 2) to (x, y) = (2, 0) evaluates (A) - 8 (C) 8

line integral over the straight line from to (B) 4 (D) 0

YEAR 2008

ONE MARKS

EE SP 10.64

X is a uniformly distributed random variable that takes values between 0 and 1. The value of E {X3} will be (A) 0 (B) 1/8 (C) 1/4 (D) 1/2

EE SP 10.65

The characteristic equation of a (3 # 3 ) matrix P is defined as a (l) = lI - P = l3 + l2 + 2l + 1 = 0 If I denotes identity matrix, then the inverse of matrix P will be (B) (P2 + P + I) (A) (P2 + P + 2I) (C) - (P2 + P + I)

EE SP 10.66

(D) - (P2 + P + 2I)

If the rank of a (5 # 6) matrix Q is 4, then which one of the following statement is correct ? (A) Q will have four linearly independent rows and four linearly independent columns (B) Q will have four linearly independent rows and five linearly independent columns


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Chapter 10

T

(C) QQ will be invertible (D) QT Q will be invertible YEAR 2008

TWO MARKS

EE SP 10.67

Consider function f (x) = (x2 - 4) 2 where x is a real number. Then the function has (A) only one minimum (B) only tow minima (C) three minima (D) three maxima

EE SP 10.68

Equation ex - 1 = 0 is required to be solved using Newton’s method with an initial guess x0 =- 1. Then, after one step of Newton’s method, estimate x1 of the solution will be given by (A) 0.71828 (B) 0.36784 (C) 0.20587 (D) 0.00000

EE SP 10.69

EE SP 10.70

nodia

A is m # n full rank matrix with m > n and I is identity matrix. Let matrix A' = (AT A) - 1 AT , Then, which one of the following statement is FALSE ? (B) (AA') 2 (A) AA'A = A (D) AA'A = A' (C) A'A = I A differential equation dx/dt = e - 2t u (t), has to be solved using trapezoidal rule of integration with a step size h = 0.01 s. Function u (t) indicates a unit step function. If x (0 -) = 0 , then value of x at t = 0.01 s will be given by (A) 0.00099 (B) 0.00495 (C) 0.0099

EE SP 10.71

(D) 0.0198

Let P be a 2 # 2 real orthogonal matrix and x is a real vector [x1, x2] T with length x = (x12 + x22) 1/2 . Then, which one of the following statements is correct ? (A) Px # x where at least one vector satisfies Px < x (B) Px # x for all vector x (C) Px $ x where at least one vector satisfies Px > x (D) No relationship can be established between x and Px YEAR 2007

EE SP 10.72

ONE MARK

x = 8x1 x2 g xn B is an n-tuple nonzero vector. The n # n matrix V = xxT (A) has rank zero (B) has rank 1 (C) is orthogonal (D) has rank n T

YEAR 2007 EE SP 10.73

TWO MARKS

1-x The differential equation dx is discretised using Euler’s numerical dt = t integration method with a time step 3 T > 0 . What is the maximum permissible


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Page 563

value of 3 T to ensure stability of the solution of the corresponding discrete time equation ? (A) 1 (B) t/2 (C) t (D) 2t EE SP 10.74

The value of C

(A) 2pi

where C # (1 dz + z2)

is the contour z -

(D) pi tan - 1 z

(C) tan z

EE SP 10.76

= 1 is

(B) p

-1

EE SP 10.75

i 2

The integral 1 2p (A) sin t cos t (C) 12 cos t

2p

#0 sin (t - t) cos tdt equals (B) 0 (D) 12 sin t

nodia A loaded dice has following probability distribution of occurrences Dice Value

1

2

3

4

5

6

Probability

1/4

1/8

1/8

1/8

1/8

1/4

If three identical dice as the above are thrown, the probability of occurrence of values 1, 5 and 6 on the three dice is (A) same as that of occurrence of 3, 4, 5 (B) same as that of occurrence of 1, 2, 5 (C) 1/128 (D) 5/8

EE SP 10.77

Let x and y be two vectors in a 3 dimensional space and < x, y > denote their dot product. Then the determinant < x, x > < x, y > det =< y, x > < y, y >G (A) is zero when x and y are linearly independent (B) is positive when x and y are linearly independent (C) is non-zero for all non-zero x and y (D) is zero only when either x or y is zero

EE SP 10.78

The linear operation L (x) is defined by the cross product L (x) = b # x , where T T b = 80 1 0B and x = 8x1 x2 x3 B are three dimensional vectors. The 3 # 3 matrix M of this operations satisfies R V Sx1 W L (x) = M Sx2 W SSx WW 3 T X Then the eigenvalues of M are (A) 0, + 1, - 1 (B) 1, - 1, 1 (C) i, - i, 1 (D) i, - i, 0


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Chapter 10

Statement for Linked Answer Question 79 and 80 Cayley-Hamilton Theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix -3 2 A == - 2 0G EE SP 10.79

EE SP 10.80

A satisfies the relation (A) A + 3I + 2A - 1 = 0 (C) (A + I) (A + 2I)

(B) A2 + 2A + 2I = 0 (D) exp (A) = 0

A9 equals (A) 511A + 510I (B) 309A + 104I (C) 154A + 155I (D) exp (9A)

nodia YEAR 2006

EE SP 10.81

EE SP 10.82

TWO MARKS

The expression V = R

#0

H

pR2 (1 - h/H) 2 dh for the volume of a cone is equal to

(A)

#0

(C)

#0 2prH (1 - r/R) dh

pR2 (1 - h/H) 2 dr

H

(B)

#0

R

(D)

#0

R

pR2 (1 - h/H) 2 dh 2 2prH`1 - r j dr R

A surface S (x, y) = 2x + 5y - 3 is integrated once over a path consisting of the points that satisfy (x + 1) 2 + (y - 1) 2 = 2 . The integral evaluates to (A) 17 2 (B) 17 2 (C) 2 /17 (D) 0

EE SP 10.83

Two fair dice are rolled and the sum r of the numbers turned up is considered (A) Pr (r > 6) = 16 (B) Pr (r/3 is an integer) =

5 6

(C) Pr (r = 8 ; r/4 is an integer) = 59 (D) Pr (r = 6 ; r/5 is an integer) = 161

Statement for Linked Answer Question 84 and 85 VT R VT R VT R S2 W S- 2 W S- 10 W P = S 1 W , Q = S- 5 W , R = S- 7 W are three vectors. SS12 WW SS 9 WW SS 3 WW X T X T X T An orthogonal set of vectors having a span that contains P, Q, Ris V R V R V R R V R V S- 4 W S 5 W S 8 W S- 6 W S 4 W (B) S 2 W S 7 W S 2 W (A) S- 3 W S- 2 W SS 4 WW SS- 11WW SS- 3 WW SS- 6 WW SS 3 WW Shop GATE Electrical in 4 Volumes by RK Kanodia at maximum discount at X T X T X T T X T X

EE SP 10.84

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EE SP 10.85


Page 565

R V R V R V R V R V R V S 6 W S- 3 W S 3 W S 4 W S 1 W S5 W (C) S 7 W S 2 W S 9 W (D) S 3 W S31W S3 W SS- 1WW SS- 2 WW SS- 4 WW SS11WW SS 3 WW SS4 WW T X T X T X T X T X T X The following vector is linearly dependent upon the solution to the previous problem V R V R S8 W S-2 W (B) S- 17 W (A) S9 W SS3 WW SS 30 WW TR V X RT VX 4 S W S13 W (C) S4 W (D) S 2 W SS5 WW SS- 3 WW T X T X

nodia YEAR 2005

EE SP 10.86

ONE MARK

In the matrix equation Px = q , which of the following is a necessary condition for the existence of at least on solution for the unknown vector x (A) Augmented matrix [Pq] must have the same rank as matrix P (B) Vector q must have only non-zero elements (C) Matrix P must be singular (D) Matrix P must be square

EE SP 10.87

If P and Q are two random events, then the following is TRUE (A) Independence of P and Q implies that probability (P + Q) = 0 (B) Probability (P , Q) $ Probability (P) + Probability (Q) (C) If P and Q are mutually exclusive, then they must be independent (D) Probability (P + Q) # Probability (P)

EE SP 10.88

If S =

#1

3 -3

x dx , then S has the value

(A) - 1 3 (C) 1 2 EE SP 10.89

(B) 1 4 (D) 1

The solution of the first order DE x' (t) =- 3x (t), x (0) = x0 is (A) x (t) = x0 e - 3t (B) x (t) = x0 e - 3 (C) x (t) = x0 e - 1/3

(D) x (t) = x0 e - 1

YEAR 2005

EE SP 10.90

TWO MARKS

V R S3 - 2 2 W For the matrix p = S0 - 2 1 W, one of the eigen values is equal to - 2 SS0 0 1 WW T is anXeigen vector ? Which of the following


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EE SP 10.91

EE SP 10.92


R V R V S3 W S- 3 W W S (A) - 2 (B) S 2 W SS 1 WW SS- 1WW TR V X RT VX S1 W S2 W (C) S- 2 W (D) S5 W SS 3 WW SS0 WW T X T X V R S1 0 - 1W If R = S2 1 - 1W, then top row of R - 1 is SS2 3 2 WW X T (A) 85 6 4B (B) 85 - 3 1B (C) 82 0 - 1B (D) 82 - 1 1/2B

nodia

A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is (A) 1/6 (B) 1/2 (C) 3/6

EE SP 10.93

EE SP 10.94

EE SP 10.95

Chapter 10

(D) 3/4

For the function f (x) = x2 e - x , the maximum occurs when x is equal to (A) 2 (B) 1 (C) 0 (D) - 1 2 y2 For the scalar field u = x + , magnitude of the gradient at the point (1, 3) is 2 3 9 13 (B) (A) 2 9 (C) 5 (D) 9 2

For the equation x'' (t) + 3x' (t) + 2x (t) = 5 ,the solution x (t) approaches which of the following values as t " 3 ? (A) 0 (B) 5 2 (C) 5

(D) 10

***********


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Page 567

SOLUTION SOL 10.1

Correct option is (B). x + 2y + 2z = b1 5x + y + 32 = b2 On eliminating y , we have 9x + 4z = 2b 2 - b1

...(i) ...(ii) {2×(ii)-(i)}

Let b1 = b2 = 1, 9x + 4z = 1 x = 1 - 4z 9 z = 1 - 9x 4 Hence these will be infinitely many solutions for any given b1 and b2 .

SOL 10.2

nodia Correct option is (A). Given the function,

f ^x h = xe-x To determine the maximum value, we equate the first derivative of function to zero, i.e. f l^x h = e-x + x ^- e-x h = e-x - xe-x = 0 or e-x ^1 - x h = 0 So, x = 1, 3 Again, we check the point of maxima by substituting the obtained values in the second derivative as f ll^x h =- e-x + xe-x - e-x -1 At x = 1 f ll^x h =- 2e-1 + ^1 h e^ h =- e-1 < 0 Hence, f ^x h is maximum at x = 1, and given as f ^1 h = ^1 h e-1 = e-1

SOL 10.3

at x = 1

Correct option is (C). Given the differential equation, d2 x =- 9x dt2 Taking the Laplace transform, s2 X ^s h - sX ^0 h - Xl^0 h =- 9X ^s h or or or or

s2 X ^s h - s ^1 h - ^1 h =- 9X ^s h X ^s h^s2 + 9h = s + 1 X ^s h = s2 + 1 = 2 s + 2 1 s +9 s +9 s +9 X ^s h = 2 s + 1 d 2 3 n s +9 3 s +9


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Chapter 10

Again, taking inverse Laplace transform, we obtain x ^ t h = cos 3t + 1 sin 3t 3 ALTERNATIVE METHOD :

Given the differential equation, d2 x + 9x = 0 dt2 Substituting dxd / D , we obtain or D2 + 9 = 0 or D = ! 3i So, the auxiliary equation is written as x = A cos 3t + iB sin 3t

nodia

...(1)

From the given values,

x ^0 h = 1 = A dx 1 and dt t = 0 = 1 = i3B & B = 3i Substituting these values in equation (1), we get x = cos 3t + 1 sin 3t 3

SOL 10.4

Correct option is (C). Given the set of points in complex plane, S = $z : z = 1. and the function, f ^z h = zz * By complex number property, we know

zz * = z 2 Thus, for the given set of points, i.e. z = 1; the value of function is f ^z h = 1 i.e. f ^z h maps S to the point (1, 0) on the plane. SOL 10.5

Correct option is (C). Given the functions,

x ^ t h ! 0 ; Tn < t < Txl and y ^ t h ! 0 ; Ty < t < Tyl z ^ t h is the convolution of x ^ t h and y ^ t h, i.e. z^t h = x^t h * y^t h So, by convolution theorem, we have z ^ t h = 0 ; t < Tx + Ty; t > Txl + Tyl or SOL 10.6

z ^ t h = 0 ; out side of Tx + Ty < T < Txl + Tyl

Correct answer is 0.5. Given the function,

g ^x h = x - 6x @


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or So, we have the graph

Page 569

g ^x h = "x ,

As this is a periodic function, the constant term will be is average value, i.e. T Average value â 0h = 1 f ^x h dx T 0 1 a 0 = 1 xdx T 0

# #

nodia 2 1 2 1 = :1 # x D = x = 0.5 2 0 2 0

Hence, the constant term is 0.5

SOL 10.7

Correct option is (B). Given the fair coin is tossed n times. Assume that Number of head appearing = x times Number of tail appearing = y times So, we have

(Given) x-y = n-3 and x+y = n Now, we consider the two cases. Case I: When x > y ; we have x-y = n-3 x+y = n 2x = 2n - 3 So, x = 2n - 3 = n - 3 2 2 As x is an integer and n is also an integer value. So, the above relation is contradictory, and therefore n-3 ! x 2 Case II: When x < y ; we have -x + y = n - 3 x+y = n y = n-3 2 Again, y is an integer and n is also an integer value. So, the above relation is contradictory. Thus, the required probability is zero.

SOL 10.8

Correct option is (B). Given function, and the circle,

Fv = yzit x2 + y2 = 1


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Chapter 10 "

Now, we have to determine the line integral of F along the circle at z = 1. From Green’s Theorem, we define line integral as " " 2Fy 2Fx ...(i) F $ dl = d n dxdy 2x 2y

#

##

"

where Fx and Fy are components of F along x and y directions, respectively, i.e. Fx = yz Fy = 0 Substituting it in equation (i), we have 2 ^yz h " " =0 G dxdy =F $ dl = 2y At z = 1 along the circle, we get

#

##

## zdxdy

nodia # F $ dl "

"

## 1 dxdy =- ## dxdy =- p ^1 h

=-

2

(area of circle)

=- p

SOL 10.9

Correct option is (B). Given the z -transform of causal signal x 6n@ as X ^z h =

1 1 1 2 -3 = 1 + 3 + c 3 m + ... 1-z z z 1 1 = 1 + 3 + 6 + ..... z z Now, the z -transform is defined as X ^z h =

3

...(i)

/ x6n@z

-n

n=0

x 61@ x 62@ x 63@ + 2 + 3 + ... z z z Comparing equations (i) and (ii), we get x 60@ = 1, x 61@ = 0 = x 60@ +

...(ii)

x 62@ = 0 , x 63@ = 1 SOL 10.10

Correct answer is 3. Given the system matrix,

R 0 1 - 1VW S A = S- 6 - 11 6W SS- 6 - 11 5WW X T The Eigen values for the system matrix is given by the roots of the equation

or

or

A - lI = 0 V R V R 0 1 - 1W Sl 0 0W S S- 6 - 11 6W - S0 l 0W = 0 SS- 6 - 11 5WW SS0 0 lWW X T T X -1 -l 1 =0 = - 6 - 11 - l 6 - 6 - 11 5 - l


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or or r

Page 571

[- l ^- 55 - 5l + 11l + l2 + 66h + 1 ^- 36 + 30 - 6lh - 1 ^66 - 66 - 6lh = 0 ^- l3 - 6l2 - 11lh - 6 - 6l + 6l = 0 l3 + 6l2 + 11l + 6 = 0

So, l =- 1, - 2 , - 3 Thus, the ratio of maximum eigenvalue to minimum eigenvalue is l max = - 3 = 3 -1 l min SOL 10.11

SOL 10.12

Correct option is (A). Corollary: Every real symmetric matrix has eigenvalues which are all real numbers.

nodia Correct answer is 0.14 . On the first dot probability \ 1 On the second dot probability \ 2

. . .

On the sixth dot probability \ 6 Total probability = 1 + 2 + 3 + 4 + 5 + 6 = 21 Probability of 3 rd dot = 3 = 1 = 0.14 21 7

SOL 10.13

SOL 10.14


f ^x h = 10 ^x - 1h2/3 Since f ^x h is a real valued function ^x - 1h can not be negative, because in that case fractional power of negative number is imaginary. So at x = 1, the value of f ^x h will be minimum. Correct option is (B). We can write or, So,

1 = cos (2np) + i sin (2np), where n is an integer 1 = ei^2nph

1i = 6ei (2np)@

i

2

1i = e-i (2np) 1i = e-2np Therefore all values of 1i are non-negative and real number. or

SOL 10.15

Correct option is (C). This is a Cauchy-Euler equation, so we substitute x = et . Then assuming x > 0 , we have t = ln x dy dt dy = dt dx dx


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Chapter 10

dy dy = b1l dx dt x dy dy = x dt dx

or or

dt 1 :t = ln x & dx = x D

2 d 2y 1 d y dt - 1 dy = x c dt2 dx m x2 dt dx2 2 d 2y 1 d y - dy = c dt m dx2 x2 dt2 d 2y d 2 y dy or x2 2 = 2 dt dx dt Substituting eq (i) and (ii) in given differential equation, we get d 2 y dy dy + -y = 0 dt dt dt2

...(i)

Now,

...(ii)

nodia d 2y -y = 0 dt2

Characteristic equation

m2 - 1 = 0 m =! 1 Roots are distinct real, so two possible solution y = et and y = e-t

SOL 10.16

Since

x = et

So

y = x or 1 are two possible solutions. x


# f# 8

0

At x =

y 2

at x =

y +1 2

y +1 2

y/2

2x - y dx p dy = ? 2 l y 2x - y = u & x = x- = u 2 2 b

dx = du 2 y/2 - y u = # =0 2 y 2 a + 1k - y =1 u = 2 2

Thus integral becomes

8

1

# ; # uduEdy 0

0

y 2 dy dv = & dy = 2dv 2 v =

at y = 0 & at y = 8 &

v =0 v =4 = Reduced form =

4

1

# ; # uduE2dv 0

0

4

1

# ; # 2uduEdv 0

0


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Page 573

Correct answer is 0.4 . 0.2

f ^x h = *0.1 0 0.2 f ^x h = *0.1 0

for x # 1 for 1 < x # 4 otherwise x = ^- 1, 1h x = ^- 4, - 1h , ^1, 4h otherwise

On solving inequalities Probability ^P h = =

# f ^x hdx 3

-3

#

1

0.5

0.2dx +

# 1

4

0.1dx

nodia = 0.2 # 0.5 + 0.1 # 3

P = 0.1 + 0.3 = 0.4

SOL 10.18


f ^x h = x3 - 3x2 - 24x + 100 on 6- 3, 3@ The given function is a polynomial and so continuous everywhere. Now we get derivative of the function to option the critical points f l^x h = 3x2 - 6x - 24 or

x2 - 2x - 8 = 0 x2 - 4x + 2x - 8 = 0 x ^x - 4h + 2 ^x - 4h = 0

^x - 4h^x + 2h = 0 So, x = 4 and x =- 2 We consider the critical point that actually fall in the given interval. So we only take x =- 2 . Now, we find the function value at the critical point x =- 2 and end points x =- 3 , x = 3 . f ^- 2h = ^- 2h3 - 3 ^- 2h2 - 24 ^- 2h + 100 f ^- 2h = 128 f ^- 3h = ^- 3h3 - 3 ^- 3h2 - 24 ^- 3h + 100 f ^- 3h = 172 f ^3h = ^3h3 - 3 ^3h2 - 24 ^3 h + 100 f ^3h = 28 We have f ^- 2h = 128 , f ^- 3h = 172 , f ^3 h = 28 From list of above values we can see that absolute minimum is 28 which occurs at x = 3 . SOL 10.19

Correct option is (C). Given the two matrices, p A => r and

q sH

p2 + q2 pr + qs B => H pr + qs r2 + s2


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Chapter 10

To determine the rank of matrix A, we obtain its equivalent matrix using the operation, a2i ! a2i - a21 a1i as a11 p q A = >0 s - r q H p If

s- r q = 0 p

ps - rq = 0 then rank of matrix A is 1, otherwise the rank is 2. Now, we have the matrix

or

p2 + q2 pr + qs B => H pr + qs r2 + s1

nodia

To determine the rank of matrix B , we obtain its equivalent matrix using the operation, a2i ! a2i - a21 a1i as a11 V R 2 2 pr + qs W Sp + q 2 W S pr qs + ^ h B = ^r2 + s2h - 2 S 0 2 W p +q W S X T ^pr + qs h2 2 2 2 If = ^ps - rq h = 0 ^r + s h - 2 p + q2 or ps - rq = 0 then rank of matrix B is 1, otherwise the rank is 2.

Thus, from the above results, we conclude that If ps - rq = 0 , then rank of matrix A and B is 1. If ps - rq ! 0 , then rank of A and B is 2. i.e. the rank of two matrices is always same. If rank of A is N then rank of B also N . SOL 10.20

Correct answer is (C). Given the velocity of particle, v = p cos pt 2 2 Since, the velocity of the particle is along x -axis. So, we have v = dx dt Therefore, the displacement of the particle from t = 0 (origin) to t = 3 is t=3 t=3 p cos p tdt x = vdt = 2 t=0 t=0 2

#

#

= 6sin p2 t @0p = 3

=- 1 or |x | = 1 i.e. the magnitude of displacement is 1. Now, we have to determine the distance covered by the particle. For calculating distance, we have to consider speed and speed can’t be negative, so the distance is given by


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x0 =

#

t=3

t=0

v dt =

Page 575

#

t=3

t=0

p cos p t dt 2 2

p cos p t - t = 3 p cos p tdt 2 2 t=0 2 t=1 2 1 t=3 = 9sin p tC - 9sin p tC 2 t=0 2 t=1

=

#

t=1

#

= 1 - ^- 1 - 1h = 3 Thus, the difference between the distance covered and magnitude of displacement is Dx = x 0 - | x | = 3 - 1 = 2 SOL 10.21

Correct answer is 0.43. Given that the life time of an electric bulb is a random variable with density, f ^x h = kx2 . where x is the lifetime measured in years. From the property of random variable, we have

nodia #

f ^x h dx = 1

3

-3

Since, the minimum and maximum lifetimes of bulbs are 1 and 2 years respectively. So, we have 1#x#2

Therefore, equation (i) becomes 2

f ^x h dx = 1

or

#

2

or

k :x D = 1 3 1 kb 8 - 1 l = 1 3 k = 3 = 0.43 7

# 1

or Thus,

SOL 10.22

1

kx2 dx = 1 3 2

Correct option is (C). Given the complex function, z2 z2 - 1 and the contour, along which integration to be performed, is C : |z - 1 | = 1 f ^z h =

Here, z - 1 = 1 is a circle with radius 1 and centre at point (1,0). So, we sketch the contour C as


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Chapter 10

Now, we have the complex function f ^z h =

z2 z2 - 1

So, the poles and zeros of the function are poles = 1, - 1 zeros = 0 , 0 Therefore, only pole (z = 1) lies within the curve C . So, the residue at z = 1 pole is 2 Residue = lim z z " 1 ^z + 1h 2 = lim z = 1 = 1 1+1 2 z"1 z + 1 Thus, the integral of complex function along the contour is obtained as

nodia # f ^z hdz C

= 2pi (sum of residues) = 2pi b 1 l = pi 2

SOL 10.23


SOL 10.24

Correct answer is 0.06005. We have the function,

f ^x h = ex - 1 So, its first derivative is obtained as f l^x h = ex By Newton-Raphson method, we have f ^xi h xi + 1 = xi f l^xi h Since, we have the initial value, x0 = 1 So, we obtain the 1st iteration as x f ^x 0h 1 = x0 - e x1 = x 0 ex f l^x 0h 1 = 1 - e -1 1 = e-1 = .3678 e Therefore, the 2nd iteration is x f ^x1h 1 = x1 - e x 2 = x1 ex f l^x1h = x1 - 1 + e-x = 0.3678 - 1 + e-0.3678 0

0

1

1

1

= .06005


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Page 577

Correct option is (B). Given, the field vector Fv = y2 xavx - yzavy - x2 avz For the line segment along x -axis, we have dlv = dxavx So, Fv : dlv = ^y2 x h^dx h Since, on x -axis y = 0 so, Fv : dlv = 0 or,

SOL 10.26

# Fv : dlv = 0

Correct option is (D). Given the equations in matrix form as 2 - 2 x1 0 > H> H = > H 1 - 1 x2 0 So, it is a homogeneous set of linear equation. It has either a trivial solution ^x1 = x2 = 0h or an infinite no. of solution. Since, for the matrix 2 -2 H A => 1 -1

nodia A =0 We have the determinant Hence, it will have multiple solutions

SOL 10.27

Correct option is (B). We know that or, So,

i = eip/2 - i = e-ip/2

(In phasor form)

- i = !ê-ip/2h = ! e-ip/4 = !=cos d p n - i sin d p nG 4 4 1/2

= cos d p n - i sin d p n ; - cos d p n + i sin d p n 4 4 4 4 = 1 - i ; -1 + i 2 2 2 This is equivalent to the given option (B) only. SOL 10.28

Correct option is (D). Given the scalar field V = 2x2 y + 3y2 z + 4z2 x Its gradient is given by dV = ^4xy + 4z2h avx + _2x2 + 6yz i avy + ^3y2 + 8zx h avz So, the curl of the gradient is obtained as avx avy avz 2 2 2 d # ^dvh = 2x 2y 2z 4xy + 4z2 2z2 + 6yz 3y2 + 8zx = avx ^6y - 6y h - avy ^8z - 8z h + avz ^4x - 4x h = 0 Note : From the properties of curl, we know that curl of gradient of any scalar


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Chapter 10

field is always zero. So, there is no need to solve the curl and gradient. SOL 10.29

Correct option is (A). Given, the PdF of random variable x as f ^x h = e-x So,

SOL 10.30

P ^x > 1h =

#e

3-x

1

0
dx = :e D = e-1 = 0.368 -1 1

Correct option is (C). Given, the equation f ^x h = x3 + 2x - 1 = 0 as initial condition is x 0 = 1.2

nodia so, from N - R method we obtain f ^xn h xn + 1 = xn f l^xn h

x 0 = 1.2 f ^x 0h = ^1.23h + 2 ^1.2h - 1 = 3.128 Also, f l^x h = 3x2 + 2 So, f l^x 0h = 3 ^1.2h2 + 2 = 6.32 Hence, 1 st iterative value is f ^x 0h x1 = x 0 f l^x 0h = 1.2 - 3.128 6.32 Here

= 0.705 SOL 10.31

Correct option is (B). Given the function y = f ^x h = 5x2 + 10x in the internal x = ^1, 2h Since, function y is continuous in the interval ^1, 2h as well as its is differentiable at each point so, from Lagranges mean value theorem there exist at least a point where f ^b h - f â h f l^c h = b-a Here, we have a=1,b=2 So, for x = a = 1, we obtain y = f â h = f ^1 h = 5 ^1 h2 + 10 ^1 h = 15 and for x = b = 2

y = f ^b h = f ^2 h = 5 ^2 h2 + 10 ^2 h = 40

Therefore, f l^c h = 40 - 15 = 25 2-1 SOL 10.32



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Page 579

Given the contour integral 2

- 4 dz +4 It has two poles given as

# zz

2

z = ! 2i Now, the contour is defined by circle z - i = 2 which is shown in the figure below

nodia

So, it can be observed that the given contour enclosed z = 2i while z =- 2i is out of the contour. So, we obtain the residue at z = 2i only as 2 ^2i h2 - 4 - 8 z 4 residue = = = = 2i z + 2i z = 2i 2i + 2i 4i Hence, contour integral is given as z2 - 4 dz = 2pi (sum of residues) z2 + 4 = 2pi ^2i h =- 4p

#

SOL 10.33

Correct option is (D). We know that the characteristic equation is given by

6A@6X@ = l 6X@

where 6A@ is the matrix as l is the scalar which gives eigen values. Now, we consider the matrix a b ^2 # 2 matrixh 6A@ = >c dH 1 For eigen value - 1 as eigen vector is > H, so, we have -1 a b 1 1 >c d H>- 1H =- 1 >- 1H or,

a - b =- 1 c-d = 1

Similarly, for a >c

1 eigen value - 2 with eigen vector > H, we obtain -2 1 b 1 =- 2 > H -2 dH>- 2H

or,

a - 2b =- 2

....(1) ....(2)

....(3)


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c - 2d = 4 Solving Eqs. (1) and (3), we obtain

Chapter 10

....(4)

a = 0, b = 1 and solving Eqs. (2) and (4), we obtain c =- 2, d = 3 Thus, the required matrix is 0 1 a b >c dH = >- 2 - 3H SOL 10.34

Correct option is (B). Probability density function of uniformly distributed variables X and Y is shown as

nodia P &[max (x, y)] < 1 0 2 Since X and Y are independent. P &[max (x, y)] < 1 0 2 PbX < 1 l 2 Similarly for Y : P bY < 1 l 2 So P &[max (x, y)] < 1 0 2

= P b X < 1 l P bY < 1 l 2 2 = shaded area = 3 4 =3 4 =3#3= 9 4 4 16

Alternate Method:

From the given data since random variables X and Y lies in the interval [- 1, 1] as from the figure X , Y lies in the region of the square ABCD . Probability for max 6X, Y @ < 1/2 : The points for max 6X, Y @ < 1/2 will be inside the region of square AEFG .


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P &max 6X, Y @ < 1 0 = Area of 4AEFG 2 Area of square ABCD 3 3 #2 2 = = 9 2#2 16

So,

SOL 10.35

Page 581

Correct option is (A). - 1 = i = cos p + i sin p 2 2

x=

p

x = ei 2

So,

xx = êi 2 h & êi 2 h = e- 2 p x

SOL 10.36

p i

p


nodia f (z) =

1 2 pj

# f (z) dz C

1 - 2 z+1 z+3

= sum of the residues of the poles which lie inside the given closed region.

C & z+1 = 1 Only pole z =- 1 inside the circle, so residue at z =- 1 is. -z + 1 f (z) = (z + 1) (z + 3) (z + 1) (- z + 1) 2 = =1 = lim 2 z "- 1 (z + 1) (z + 3) 1 So f (z) dz = 1 2 pj C

#

SOL 10.37

Correct option is (D). t dx + x = t dt dx + x = 1 t dt dx + Px = Q (General form) dt IF = e # = e = e lnt = t

Integrating factor, Solution has the form

Pdt

1 # dt t

# ^Q # IF hdt + C x # t = # (1) (t) dt + C

x # IF =

2

xt = t + C 2 Taking the initial condition

C =0 2 xt = t & x = t 2 2

So, SOL 10.38

x (1) = 0.5 0.5 = 1 + C 2

Correct option is (B). Characteristic equation. A - lI = 0


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Chapter 10

-5 - l -3 =0 2 -l 5l + l2 + 6 = 0 l2 + 5l + 6 = 0 Since characteristic equation satisfies its own matrix, so A2 + 5A + 6 = 0 & A2 =- 5A - 6I Multiplying with A A3 + 5A2 + 6A = 0 A3 + 5 (- 5A - 6I) + 6A = 0 A3 = 19A + 30I SOL 10.39


&

f (x) = x3 - 9x2 + 24x + 5 df (x) = 3x2 - 18x + 24 = 0 dx df (x) = x2 - 6x + 8 = 0 dx

x = 4, x = 2

d 2 f (x) = 6x - 18 dx 2 d 2 f (x) For x = 2, = 12 - 18 =- 6 < 0 dx2 So at x = 2, f (x) will be maximum f (x)

SOL 10.40

max

= (2) 3 - 9 (2) 2 + 24 (2) + 5 = 8 - 36 + 48 + 5 = 25

Correct option is (C). Probability of appearing a head is 1/2. If the number of required tosses is odd, we have following sequence of events. H, TTH, TTTTH, ........... 1

3 5 P = 1 + b 1 l + b 1 l + ..... = 2 1 = 2 3 2 2 2 1- 4

Probability SOL 10.41

Correct option is (A). Divergence of A in spherical coordinates is given as d:A = 12 2 (r 2 Ar ) r 2r = 12 2 (krn + 2) = k2 (n + 2) rn + 1 r 2r r n-1 = k (n + 2) r = 0 (given) n + 2 = 0 or n =- 2

SOL 10.42

Correct option is (D). d 2 y (t) 2dy (t) + + y (t) = d (t) dt dt 2 Taking Laplace transform with initial conditions


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dy 2 ;s Y (s) - sy (0) - dt &

Page 583

E + 2 [sy (s) - y (0)] + Y (s) = 1

t=0

2 6s Y (s) + 2s - 0@ + 2 6sY (s) + 2@ + Y (s) = 1

Y (s) [s2 + 2s + 1] = 1 - 2s - 4 Y (s) = 2- 2s - 3 s + 2s + 1 We know If,

y (t) dy (t) dt

then,

L

Y (s)

L

sY (s) - y (0)

(- 2s - 3) s +2 (s2 + 2s + 1) 2 2 = - 2s - 32 s + 2s + 4s + 2 (s + 2s + 1) 1 1 = 1 + sY (s) - y (0) = s + 2 2 = s + 1 2 + s + 1 (s + 1) 2 (s + 1) (s + 1) (s + 1) 2 Taking inverse Laplace transform dy (t) = e-t u (t) + te-t u (t) dt dy At t = 0+ , = e0 + 0 = 1 dt t = 0 So,

sY (s) - y (0) =

nodia +

SOL 10.43


x3+x2+x+1 = 0 x 2 (x + 1) + (x - 1) = 0

(x + 1) (x 2 + 1) = 0 x + 1 = 0 & x =- 1

or

x2 + 1 = 0 & x =- j, j x =- 1, - j, j

and

SOL 10.44


Z is Z = 0 where q is around 45c or so. Thus

Z = Z 45c where Z < 1 1 = 1 - 45c Y = 1 = Z Z Z 45c Y > 1 [a Z < 1]


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Chapter 10

So Y will be out of unity circle. SOL 10.45

Correct option is (A). dy = e-3x dx

nodia dy = e-3x dx

by integrating, we get

y =- 1 e-3x + K , where K is constant. 3

SOL 10.46


f 1 = 10x2 sin x1 - 0.8 f 2 = 10x 22 - 10x2 cos x1 - 0.6 Jacobian matrix is given by V R S2f 1 2f 1 W 10x2 cos x1 10 sin x1 2x 2x J = SS2f 21 2f 22 WW = > 10x2 sin x1 20x2 - 10 cos x1H S2x1 2x2 W X T 10 0 For x1 = 0, x2 = 1, J => 0 10H

SOL 10.47

Correct option is (C). f (x) = 2x - x 2 + 3 f l (x) = 2 - 2x = 0 x =1 f m (x) =- 2 f m (x) is negative for x = 1, so the function has a maxima at x = 1.

SOL 10.48

Correct option is (A). Let a signal p (x) is uniformly distributed between limits - a to + a .

Variance

sp =

#

a

-a

x 2 p (x) dx


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=

#

a

-a

Page 585

x2 : 1 dx 2a

3 2 3 a = 1 :x D = 2a = a 2a 3 -a 6 3 It means square value is equal to its variance 2 2 = sp = a p rms 3 p rms = a 3

SOL 10.49

Correct option is (D). We know that matrix A is equal to product of lower triangular matrix L and upper triangular matrix U .

nodia A = 6L@6U @ only option (D) satisfies the above relation.

SOL 10.50


Let the given two vectors are

X1 = [1, 1, 1] X2 = [1, a, a 2] Dot product of the vectors

Where

R1V S W X 1 $ X 2 = X1 X 2T = 81 1 1BS a W = 1 + a + a2 SSa 2WW T X a =- 1 + j 3 = 1 - 2p/3 2 2

so, X1, X2 are orthogonal 1 + a + a2 = 0 Note: We can see that X1, X2 are not orthonormal as their magnitude is ! 1 SOL 10.51

Correct option is (B). P =

#0

1

xex dx

= 6x # e = 6xe

x

dx @0 - # 1 : d

@0 - #0

x 1

1

0

1

dx

(x) # ex dx D dx

(1) ex dx

= (e1 - 0) - 6e

@0

x 1


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Chapter 10

= e1 - [e1 - e0] = 1 SOL 10.52

Correct option is (A). t Radial vector r = xti + ytj + zk Divergence = 4$ r t : _xti + ytj + zk ti = c 2 ti + 2 tj + 2 k 2x 2y 2z m 2y 2z = 2x + + 2x 2y 2z = 1+1+1 = 3

SOL 10.53

SOL 10.54

Correct option is (C). No of white balls = 4 , no of red balls = 3 If first removed ball is white then remaining no of balls = 6 (3 white, 3 red) we have 6 balls, one ball can be choose in 6 C1 ways, since there are three red balls so probability that the second ball is red is 6 P = 3 C1 = 3 = 1 6 2 C1

nodia Correct option is (B). Let eigen vector

X = 8x1 x2 x 3BT Eigen vector corresponding to l1 = 1 8A R0 S S0 SS0 T

I BX = 0 Rx V R0V S 1W S W Sx2W = S0W SSx WW SS0WW 3 T X T X x2 = 0 x2 + 2x 3 = 0 & x 3 = 0 (not given in the option) Eigen vector corresponding to l2 = 2 8A - l2 I B X = 0 R- 1 1 0V Rx V R0V S W S 1W S W S 0 0 2W Sx2W = S0W SS 0 0 1WW SSx WW SS0WW 3 T X T X T X - x1 + x 2 = 0 2x 3 = 0 & x 3 = 0 (not given in options.) Eigen vector corresponding to l3 = 3 - l1 1 0VW 1 2W 0 2WW X

8A - l3 I B X = 0 R- 2 1 0V Rx V R0V S W S 1W S W S 0 - 1 2W Sx2W = S0W SS 0 0 0WW SSx WW SS0WW 3 T X T X T X - 2x1 + x2 = 0

- x 2 + 2x 3 = 0 Put x1 = 1, x2 = 2 and x 3 = 1


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So Eigen vector

SOL 10.55

Page 587

Rx V R1V S 1W S W X = Sx2W = S2W = 81 2 1BT SSx WW SS1WW 3 T X T X

Correct option is (B). d2 x + 6 dx + 8x = 0 dt dt2 Taking Laplace transform (with initial condition) on both sides s2 X (s) - sx (0) - x' (0) + 6 [sX (s) - x (0)] + 8X (s) = 0 s2 X (s) - s (1) - 0 + 6 [sX (s) - 1] + 8X (s) = 0 X (s) [s2 + 6s + 8] - s - 6 = 0

nodia X (s) =

(s + 6) (s2 + 6s + 8)

By partial fraction

2 - 1 s+2 s+4 Taking inverse Laplace transform X (s) =

x (t) = (2e- 2t - e- 4t)

SOL 10.56

Correct option is (C). Set of equations

.....(1) x1 + 2x2 + x 3 + 4x 4 = 2 .....(2) 3x1 + 6x2 + 3x 3 + 12x 4 = 6 or 3 (x1 + 2x2 + x 3 + 4x 4) = 3 # 2 Equation (2) is same as equation(1) except a constant multiplying factor of 3. So infinite (multiple) no. of non-trivial solution exists.

SOL 10.57

Correct option is (D). Function f (t)= sin t = sin ct has a maxima at t = 0 as shown below t


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Chapter 10

Correct option is (C). a b A => H c d Trace of a square matrix is sum of its diagonal entries Let the matrix is

Trace A = a + d =- 2 Determinent Eigenvalue

ad - bc =- 35 A - lI = 0 a-l b =0 c d-l

(a - l) (d - l) - bc = 0 l - (a + d) l + (ad - bc) = 0 2

l2 - (- 2) l + (- 35) = 0 l2 + 2l - 35 = 0

nodia (l - 5) (l + 7) = 0 l1, l2 = 5, - 7

SOL 10.59

Correct option is (A). Given constraints x > y2 and y > x2

Limit of y : y = 0 to y = 1 Limit of x : x = y2 to x2 = y & x = So volume under f (x, y) V =

y=1

x= y

#y = 0 #x = y

2

y

f (x, y) dx dy

SOL 10.60

Correct option is (B). No of events of at least two people in the room being born on same date = n C2 three people in the room being born on same date = n C 3 Similarly four for people = n C 4 n n n n Probability of the event, 0.5 $ C2 $ C 3 $ C 4 g Cn & N = 7 N

SOL 10.61

Option ( ) is correct. Assume a Cubic polynomial with real Coefficients P (x) = a 0 x3 + a1 x3 + a2 x + a 3

a 0, a1, a2, a 3 are real


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Page 589

P' (x) = 3a 0 x2 + 2a1 x + a2 P'' (x) = 6a 0 x + 2a1 P''' (x) = 6a 0 Piv (x) = 0 SOL 10.62

Correct option is (D). An iterative sequence in Newton-Raphson’s method is obtained by following expression f (xk ) xk + 1 = xk f' (xk )

nodia So

So

SOL 10.63

f (x) = x2 - 117 f' (x) = 2x f (xk ) = x k2 - 117 f' (xk ) = 2xk = 2 # 117 2 xk + 1 = xk - x k - 117 = xk - 1 :xk + 117 D 2xk 2 xk

Correct option is (D). Equation of straight line y - 2 = 0 - 2 (x - 0) 2-0 y - 2 =- x F $ dl = [(x2 + xy) at x + (y2 + xy) at y] [dxat x + dyat y + dzat z] = (x2 + xy) dx + (y2 + xy) dy Limit of x : 0 to 2 Limit of y : 2 to 0

# F $ dl Line

#0

2

(x2 + xy) dx + y - 2 =- x =

#2

0

(y2 + xy) dy

dy =- dx So

SOL 10.64

# F $ dl

=

#0

2

[x2 + x (2 - x)] dx +

=

#0

2

2xdx +

#2

0

#2

0 2

y + (2 - y) y dy

2 2 y2 0 2y dy = 2 :x D + 2 ; E = 4 - 4 = 0 2 0 2 2

Correct option is (C). X is uniformly distributed between 0 and 1 So probability density function 1, 0 < x < 1 fX (X) = ) 0, otherwise 1 3 So, E {X } = # X3 fX (X) dx = 0

#0

1

X3 (1) dx

4 1 = :X D = 1 4 0 4

SOL 10.65

Correct option is (D). According to CAYLEY-HAMILTON Theorem every non-singular square matrix satisfies its own characteristic equation. Characteristic equation


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Chapter 10

a (l) = lI - P = l3 + l2 + 2l + 1 = 0 Matrix P satisfies above equation P 3 + P 2 + 2P + I = 0 I =- (P3 + P2 + 2P) Multiply both sides by P- 1 P- 1 =- (P2 + P + 2I) SOL 10.66

SOL 10.67

Correct option is (A). Rank of a matrix is no. of linearly independent rows and columns of the matrix. Here Rank r (Q) = 4 So Q will have 4 linearly independent rows and flour independent columns.

nodia Correct option is (B). Given function

f (x) = (x2 - 4) 2 f' (x) = 2 (x2 - 4) 2x To obtain minima and maxima f' (x) = 0 4x (x - 4) = 0 x = 0, x2 - 4 = 0 & x = ! 2 x = 0, + 2, - 2 2

So,

f'' (x) = 4x (2x) + 4 (x2 - 4) = 12x2 - 16

x = 0, f'' (0) = 12 (0) 2 - 16 =- 16 < 0 x =+ 2, f'' (2) = 12 (2) 2 - 16 = 32 > 0 x =- 2, f'' (- 2) = 12 (- 2) 2 - 16 = 32 > 0 So f (x) has only two minima For

SOL 10.68

(Maxima) (Minima) (Minima)

Correct option is (A). An iterative sequence in Newton-Raphson method can obtain by following expression f (xn) xn + 1 = xn f' (xn) We have to calculate x1 , so n = 0 x1 = x 0 -

f (x 0) , Given x 0 =- 1 f' (x 0)

f (x 0) = ex - 1 = e- 1 - 1 =- 0.63212 f' (x 0) = ex = e- 1 = 0.36787 0

0

x1 =- 1 -

So,

(- 0.63212) (0.36787)

=- 1 + 1.71832 = 0.71832 SOL 10.69

Correct option is (D). A' = (AT A) - 1 AT = A- 1 (AT ) - 1 AT = A- 1 I


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Page 591

Put A' = A- 1 I in all option. AA'A = A

option (A)

AA- 1 A = A A =A (AA') 2 = I (AA- 1 I) 2 = I

option (B)

(true)

(I) 2 = I A'A = I

option (C)

(true)

A- 1 IA = I I =I AA'A = A' AA- 1 IA = A = Y A'

option (D)

SOL 10.70

(true)

nodia

(false)


dx = e- 2t u (t) dt

x = # e- 2t u (t) dt =

1

# e- 2t dt

0

=

1

# f (t) dt ,

0

t = .01 s

From trapezoid rule

#

t0

f (t) dt = h 6f (0) + f (.01)@ 2 1 e0 + e- .02@, h = .01 #0 f (t) dt = .01 2 6

t 0 + nh

= .0099 SOL 10.71

Correct option is (B). P is an orthogonal matrix So PPT = I cos q - sin q Let assume P => sin q cos q H cos q - sin q PX = > x x T sin q cos q H8 1 2B cos q - sin q x1 x1 cos q - x2 sin q => => H > H sin q cos q x2 x1 sin q + x2 cos qH PX PX

SOL 10.72

=

(x1 cos q - x2 sin q) 2 + (x1 sin q + x2 cos q) 2

=

x 12 + x 22

= X

Correct option is (D). x = 8x1 x2 g xnBT V = xxT


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R V Sx1W Sx2W =S W ShW SxnW T X

So rank of V is n . SOL 10.73


SOL 10.74


Chapter 10

R V Sx1W Sx2W ShW S W SxnW T X

dz = # dz 2 + ( z i ) (z - i) 1 z + C C Contour z- i = 1 2 P(0, 1) lies inside the circle z - i = 1 and P (0, 1) does not lie. 2 So by Cauchy’s integral formula 1 (z - i) # dz 2 = 2pi lim z"i + ( z i ) (z - i) 1 z + C = 2pi lim 1 = 2pi # 1 = p 2i z"i z + i Given

SOL 10.75

SOL 10.76

#

nodia Correct option is ( ).

Correct option is (C). Probability of occurrence of values 1,5 and 6 on the three dice is P (1, 5, 6) = P (1) P (5) P (6) = 1#1#1 = 1 128 8 4 4 In option (A) P (3, 4, 5) = P (3) P (4) P (5) =1#1#1 = 1 8 8 8 512 In option (B) P (1, 2, 5) = P (1) P (2) P (5) = 1#1#1 = 1 8 8 256 4

SOL 10.77

Correct option is (D). x$x x$y det >y $ x y $ yH = (x : x) (y : y) - (x : y) (y : x) = 0 only when x or y is zero

SOL 10.78


SOL 10.79

Correct option is (C). For characteristic equation -3 - l 2 > - 1 0 - lH = 0 or

(- 3 - l) (- l) + 2 = 0


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Page 593

(l + 1) (l + 2) = 0 According to Cayley-Hamiliton theorem (A + I) (A + 2I) = 0 SOL 10.80

Correct option is (A). According to Cayley-Hamiliton theorem or or

(A + I) (A + 2I) = 0 A 2 + 3A + 2 I = 0 A2 =- (3A + 2I) A 4 = (3A + 2I) 2 = (9A2 + 12A + 4I)

or

= 9 (- 3A - 2I) + 12A + 4I =- 15A - 14I A = (- 15A - 14I) 2 = 225A2 + 420A + 196 = 225 (- 3A - 2I) + 420A + 196I 8

nodia =- 255A - 254I A =- 255A2 - 254A =- 255 (- 3A - 2I) - 254A = 511A + 510I 9

SOL 10.81

Correct option is (D). Volume of the cone

V =

#0

H

2 pR2 b1 - h l dh H

Solving the above integral V = 1 pR 2 H 3 Solve all integrals given in option only for option (D) R #0 2prH a1 - Rr k2 dr = 13 pR2 H SOL 10.82


SOL 10.83

Correct option is (C). By throwing dice twice 6 # 6 = 36 possibilities will occur. Out of these sample space consist of sum 4, 8 and 12 because r/4 is an integer. This can occur in following way : (1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2) and (6, 6) Sample Space =9 Favourable space is coming out of 8 =5 Probability of coming out 8 =5 9

SOL 10.84


SOL 10.85


SOL 10.86



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Chapter 10

Matrix equation PX = q has a unique solution if Where

r (P) = r (r) r (P) " rank of matrix P r (r) " rank of augmented matrix [P] r = 8P : qB

SOL 10.87

Correct option is (D). for two random events conditional probability is given by probability (P + Q) = probability (P) probability (Q) probability (P + Q) #1 probability (Q) = probability (P) so

SOL 10.88

probability (P + Q) # probability (P)


S =

# 3 x- 3 dx

1

-2 3 = :x D = 1 -2 1 2

SOL 10.89

Correct option is (A). We have or

A.E.

Thus solution is From x (0) = x 0 we get Thus

xo(t) =- 3x (t) xo(t) + 3x (t) = 0 D+3 = 0 x (t) = C1 e- 3t C1 = x 0 x (t) = x 0 e- 3t

SOL 10.90

Correct option is (D). For eigen value l =- 2 R3 - (- 2) V Rx V R0V -2 2 S W S 1W S W - 2 - (- 2) 0 1 S W Sx2W = S0W SS 0 0 1 - (- 2)WW SSx 3WW SS0WW T R5 - 2 2VX RTx VX RT0VX S W S 1W S W S0 0 1W Sx2W = S0W SS0 0 1WW SSx WW SS0WW 3 T X T X T X 5x1 - 2x2 + x 3 = 0

SOL 10.91

Correct option is (B). C11 = 2 - (- 3) = 5 C21 =- (0 - (- 3)) =- 3 C 31 = (- (- 1)) = 1 R = (1) C11 + 2C21 + 2C 31 = 5-6+2 = 1

SOL 10.92



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Page 595

If the toss produces head, then for exactly two head in three tosses three tosses there must produce one head in next two tosses. The probability of one head in two tosses will be 1/2. SOL 10.93

Correct option is (A). f (x) = x2 e- x or f' (x) = 2xe- x - x2 e- x = xe- x (2 - x) f'' (x) = (x2 - 4x + 2) e- x Now for maxima and minima, f' (x) = 0 We have

xe- x (2 - x) = 0 x = 0, 2

or

nodia

at x = 0 f'' (0) = 1 (+ ve) at x = 2 f'' (2) =- 2e- 2 (- ve) Now f'' (0) = 1 and f'' (2) =- 2e- 2 < 0 . Thus x = 2 is point of maxima

SOL 10.94


4 u = cti 2 + tj 2 m u 2x 2y = ti2u + tj2u 2x 2y = xti + 2 ytj 3

At (1, 3) magnitude is 4u = SOL 10.95

x2 + b 2 y l = 1 + 4 = 3 2

5

Correct option is (B). d2 x + 3dx + 2x (t) = 5 dt dt2 Taking Laplace transform on both sides of above equation. s2 X (s) + 3sX (s) + 2X (s) = 5 s 5 X (s) = 2 s (s + 3s + 2) From final value theorem lim x (t) = lim X (s)

t"3

s"0

= lim s s"0

5 =5 2 s (s + 3s + 2) 2

***********


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General Aptitudes

Chapter 11

CHAPTER 11 GENERAL APTITUDES


EE SP 11.2

ONE MARK

Which of the following options is the closest in meaning to the phrase underlined in the sentence below ? It is fascinating to see life forms cope with varied environmental conditions. (A) adopt (B) adapt to (C) adept in (D) accept with

nodia

Choose the most appropriate word from the options given below to complete the following sentence. He could not understand the judges awarding her the first prize, because he thought that her performance was quite __________. (A) superb (B) medium (C) mediocre

EE SP 11.3

EE SP 11.4

EE SP 11.5

(D) exhilarating

In a press meet on the recent scam, the minister said, “The buck stops here”. What did the minister convey by the statement ? (A) He wants all the money (B) He will return the money (C) He will assume final responsibility (D) He will resist all enquiries If ^z + 1/z h2 = 98 , compute ^z2 + 1/z2h.

The roots of ax2 + bx + c = 0 are real and positive. a , b and c are real. Then ax2 + b x + c = 0 has (A) no roots (B) 2 real roots (C) 3 real roots (D) 4 real roots YEAR 2014 EE01

EE SP 11.6

TWO MARKS

The Palghat Gap (or Palakkad Gap), a region about 30 km wide in the southern part of the Western Ghats in India, is lower than the hilly terrain to its north and south. The exact reasons for the formation of this gap are not clear. It results in the neighbouring regions of Tamil Nadu getting more rainfall from the South West mansoon and the neighbouring regions of Kerala having higher summer temperatures. What can be inferred from this passage ? (A) The Palghat gap is caused by high rainfall and high temperatures in southern Tamil Nadu nad Kerala (B) The regions in Tamil Nadu and Kerala that are near the Palghat Gap are low-lying


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General Aptitudes

Page 597

(C) The low terrian of the Palghat Gap has a significant impact on weather patterns in neighbouring parts of Tamil Nadu and Kerala. (D) Higher summer temperature result in higher rainfall near the Palghat Gap area. EE SP 11.7

Geneticists say that they are very close to confirming the genetic roots of psychiatric illnesses such as depression and schizophrenia, and consequently, that doctors will be able to eradicate these diseases through early identification and gene therapy. On which of the following assumptions does the statement above rely ? (A) Strategies are now available for eliminating psychiatric illnesses (B) Certain psychiatric illnesses have a genetic basis (C) All human diseases can be traced back to genes and how they are expressed

nodia

(D) In the future, genetics will become the only relevant field for identifying psychiatric illnesses

EE SP 11.8

EE SP 11.9

Round-trip tickets to a tourist destination are eligible for a discount of 10% on the total fare. In addition, groups of 4 or more get a discount of 5% on the total fare. If the one way single person fare is Rs 100, a group of 5 tourists purchasing round-trip tickets will be charged Rs _________. In a survey, 300 respondents were asked whether they own a vehicle or not. If yes, they were further asked to mention whether they own a car or scooter or both. Their responses are tabulated below. What percent of respondents do not own a scooter ?

Own vehicle

Men

Women

Car

40

34

Scooter

30

20

Both

60

46

20

50

Do not own vehicle EE SP 11.10

When a point inside of a tetrahedron (a solid with four triangular surfaces) is connected by straight lines to its corners, how many (new) internal planes are created with these lines ? _____________ YEAR 2014 EE02

EE SP 11.11

ONE MARK

Choose the most appropriate phrase from the options given below to complete the following sentence. India is a post-colonial country because (A) it was a former British colony (B) Indian Information Technology professionals have colonized the world (C) India does not follow any colonial practices (D) India has helped other countries gain freedom


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General Aptitudes

Chapter 11

EE SP 11.12

Who ______ was coming to see us this evening ? (A) you said (B) did you say (C) did you say that (D) had you said

EE SP 11.13

Match the columns Column 1

Column 2

1.

eradicate

P.

misrepresent

2.

distort

Q.

soak completely

3.

saturate

R.

use

4.

utilize

S.

destroy utterly

nodia (A) 1 : S, 2 : P, 3 : Q, 4 : R (B) 1 : P, 2 : Q, 3 : R, 4 : S

(C) 1 : Q, 2 : R, 3 : S, 4 : P (D) 1 : S, 2 : P, 3 : R, 4 : Q

EE SP 11.14

EE SP 11.15

What is the average of all multiplies of 10 from 2 to 198 ? (A) 90 (B) 100 (C) 110 (D) 120

The value of (A) 3.464 (C) 4.000

12 + 12 + 12 + ... is (B) 3.932 (D) 4.444

YEAR 2014 EE02

TWO MARKS

EE SP 11.16

The old city of Koenigsberg, which had a German majority population before World War 2, is now called Kaliningrad. After the events of the war, Kaliningrad is now a Russian territory and has a predominantly Russian population. It is bordered by the Baltic Sea on the north and the countries of Poland to the south and west and Lithuania to the east respectively. Which of the statement below can be inferred from this passage ? (A) Kaliningrad was historically Russian in its ethnic make up (B) Kaliningrad is a part of Russia despite it not being contiguous with the rest of Russia (C) Koenigsberg was renamed Kaliningrad, as that was its original Russain name (D) Poland and Lithuania are on the route from Kaliningrad to the rest of Russia

EE SP 11.17

The number of people diagnosed with dengue fever (contracted from the bite of a mosquito) in north India is twice the number diagnosed last year. Municipal authorities have concluded that measures to control the mosquito population


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General Aptitudes

Page 599

have failed in this region. Which one of the following statements, if true, does not contradict this conclusion ? (A) A high proportion of the affected population has returned from neighbouring countries where dengue is prevalent (B) More cases of dengue are now reported because of an increase in the Municipal Office’s administrative efficiency (C) Many more cases of dengue are being diagnosed this year since the introduction of a new and effective diagnostic test (D) The number of people with malarial fever (also contracted from mosquito bites) has increased this year EE SP 11.18

EE SP 11.19

EE SP 11.20

nodia

If x is real and x2 - 2x + 3 = 11, then possible values of - x3 + x2 - x include (A) 2, 4 (B) 2, 14 (C) 4, 52 (D) 14, 52 The ratio of male to female students in a college for five years is plotted in the following line graph. If the number of female students doubled in 2009, by what percent did the number of male students increase in 2009 ?

At what time between 6 a.m. and 7 a.m. will the minute hand and hour hand of a clock make an angle closest to 60c ? (B) 6 : 27 a.m. (A) 6 : 22 a.m. (C) 6 : 38 a.m. (D) 6 : 45 a.m. YEAR 2014 EE03

EE SP 11.21

ONE MARK

While trying to collect an envelope from under the table , Mr. X fell down and II

I

III

was losing consciousness IV

Which one of the above underlined parts of the sentence is NOT appropriate ? (A) I (B) II (C) III (D) IV


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General Aptitudes

Chapter 11

If she ____ how to calibrate the instrument, she _____ done the experiment. (A) knows, will have (B) knew, had (C) had known, could have (D) should have known, would have

EE SP 11.23

Choose the word that is opposite in meaning to the word “coherent”. (A) sticky (B) well-connected (C) rambling (D) friendly

EE SP 11.24

EE SP 11.25

nodia Which number does not belong in the series below ? 2, 5, 10, 17, 26, 37, 50, 64 (A) 17 (B) 37 (C) 64 (D) 26

The table below has question-wise data on the performance of students in an examination. The marks for each question are also listed. There is no negative or partial marking in the examination. Q. No.

Marks

Answered Correctly Answered Wrongly

Not Attempted

1

2

21

17

6

2

3

15

27

2

3

2

23

18

3

What is the average of the marks obtained by the class in the examination ? (A) 1.34 (B) 1.74 (C) 3.02 (D) 3.91 YEAR 2014 EE03 EE SP 11.26

TWO MARKS

A dance programme is scheduled for 10.00 a.m. Some students are participating in the programme and they need to come an hour earlier than the start of the event. These students should be accompanied by a parent. Other students and parents should come in time for the programme. The instruction you think that is appropriate for this is (A) Students should come at 9.00 a.m. and parents should come at 10.00 a.m. (B) Participating students should come at 9.00 a.m. accompanied by a parent, and other parents and students should come by 10.00 a.m. (C) Students who are not participating should come by 10.00 a.m. and they should not bring their parents. Participating students should come at 9.00 a.m. (D) Participating students should come before 9.00 a.m. Parents who accompany them should come at 9.00 a.m. All others should come at 10.00 a.m.


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General Aptitudes

Page 601

By the beginning of the 20 th century, several hypotheses were being proposed, suggesting a paradigm shift in our understanding of the universe. However, the clinching evidence was provided by experimental measurements of the position of a star which was directly behind our sun. Which of the following inference(s) may be drawn from the above passage ? (i) Our understanding of the universe changes based on the positions of stars (ii) Paradigm shifts usually occur at the beginning of centuries (iii) Stars are important objects in the universe (iv) Experimental evidence was important in confirming this paradigm shift (A) (i), (ii) and (iv) (C) (i) and (iv)

EE SP 11.28

(B) (iii) only (D) (iv) only

nodia

The Gross Domestic Product (GDP) in Rupees grew at 7% during 2012-2013. For international comparison, the GDP is compared in US Dollars (USD) after conversion based on the market exchange rate. During the period 2012-2013 the exchange rate for the USD increased from Rs. 50/ USD to Rs. 60/USD. India’a GDP in USD during the period 2012-2013. (A) increased by 5% (B) decreased by 13% (C) decreased by 20%

(D) decreased by 11%

EE SP 11.29

The ratio of male to female students in a college for five years is plotted in the following line graph. If the number of female students in 2011 and 2012 is equal, what is the ratio of male students in 2012 to male students in 2011 ?

(A) 1 : 1 (C) 1.5 : 1 EE SP 11.30

(B) 2 : 1 (D) 2.5 : 1

Consider the equation ^7526h8 - ^Y h8 = ^4364h8 , where ^X hN stands for X to the base N . Find Y . (B) 1737 (A) 1634 (C) 3142 (D) 3162


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General Aptitudes

Chapter 11

YEAR 2013 EE SP 11.31

ONE MARK

Choose the grammatically CORRECT sentence: (A) Two and two add four (B) Two and two become four (C) Two and two are four (D) Two and two make four

EE SP 11.32

Statement: You can always give me a ring whenever you need. Which one of the following is the best inference from the above statement? (A) Because I have a nice caller tune. (B) Because I have a better telephone facility (C) Because a friend in need is a friend indeed

nodia

(D) Because you need not pay towards the telephone bills when you give me a ring

EE SP 11.33

In the summer of 2012, in New Delhi, the mean temperature of Monday to Wednesday was 41°C and of Tuesday to Thursday was 43cC . If the temperature on Thursday was 15% higher than that of Monday, then the temperature in cC on Thursday was (A) 40 (B) 43 (C) 46

(D) 49

EE SP 11.34

Complete the sentence: Dare .................. mistakes. (A) commit (B) to commit (C) committed (D) committing

EE SP 11.35

They were requested not to quarrel with others. Which one of the following options is the closest in meaning to the word quarrel? (A) make out (B) call out (C) dig out (D) fall out

SOL 11.1

Correct option is (D). They were requested not to quarrel with others. Quarrel has a similar meaning to ‘fall out’ YEAR 2013

TWO MARKS

EE SP 11.36

A car travels 8 km in the first quarter of an hour, 6 km in the second quarter and 16 km in the third quarter. The average speed of the car in km per hour over the entire journey is (A) 30 (B) 36 (C) 40 (D) 24

EE SP 11.37

Find the sum to n terms of the series 10 + 84 + 734 + ...


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General Aptitudes

(A) (C) EE SP 11.38

9 ^9n + 1h +1 10

9 ^9n - 1h +n 8

Page 603

(B) (D)

9 ^9n - 1h +1 8

9 ^9n - 1h + n2 8

Statement: There were different streams of freedom movements in colonial India carried out by the moderates, liberals, radicals, socialists, and so on. Which one of the following is the best inference from the above statement? (A) The emergence of nationalism in colonial India led to our Independence (B) Nationalism in India emerged in the context of colonialism (C) Nationalism in India is homogeneous (D) Nationalism in India is heterogeneous

EE SP 11.39

EE SP 11.40

nodia

The set of values of p for which the roots of the equation 3x2 + 2x + p ^p - 1h = 0 are of opposite sign is (A) ^- 3, 0h (B) ^0, 1h (C) ^1, 3h (D) ^0, 3h What is the chance that a leap year, selected at random, will contain 53 Sundays? (A) 2/7 (B) 3/7 (C) 1/7 (D) 5/7 2012

ONE MARK

EE SP 11.41

If (1.001) 1259 = 3.52 and (1.001) 2062 = 7.85, then (1.001) 3321 (B) 4.33 (A) 2.23 (D) 27.64 (C) 11.37

EE SP 11.42

Choose the most appropriate alternate from the options given below to complete the following sentence : If the tired soldier wanted to lie down, he..................the mattress out on the balcony. (A) should take (B) shall take (C) should have taken (D) will have taken

EE SP 11.43

Choose the most appropriate word from the options given below to complete the following sentence : Give the seriousness of the situation that he had to face, his........was impressive. (A) beggary (B) nomenclature (C) jealousy (D) nonchalance

EE SP 11.44

Which one of the following options is the closest in meaning to the word given below ? Latitude (A) Eligibility (B) Freedom (C) Coercion (D) Meticulousness


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General Aptitudes

Chapter 11

One of the parts (A, B, C, D) in the sentence given below contains an ERROR. Which one of the following is INCORRECT ? I requested that he should be given the driving test today instead of tomorrow. (A) requested that (B) should be given (C) the driving test

(D) instead of tomorrow

2012 EE SP 11.46

TWO MARKS

One of the legacies of the Roman legions was discipline. In the legious, military law prevailed and discipline was brutal. Discipline on the battlefield kept units obedient, intact and fighting, even when the odds and conditions were against them. Which one of the following statements best sums up the meaning of the above passage ? (A) Through regimentation was the main reason for the efficiency of the Roman legions even in adverse circumstances.

nodia (B) The legions were treated inhumanly as if the men were animals (C) Disciplines was the armies inheritance from their seniors

(D) The harsh discipline to which the legions were subjected to led to the odds and conditions being against them.

EE SP 11.47

Raju has 14 currency notes in his pocket consisting of only Rs. 20 notes and Rs. 10 notes. The total money values of the notes is Rs. 230. The number of Rs. 10 notes that Raju has is (A) 5 (B) 6 (C) 9 (D) 10

EE SP 11.48

There are eight bags of rice looking alike, seven of which have equal weight and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance, the minimum number of weighings required to identify the heavier bag is (A) 2 (B) 3 (C) 4 (D) 8

EE SP 11.49

The data given in the following table summarizes the monthly budget of an average household. Category

Amount (Rs.)

Food

4000

Clothing

1200

Rent

2000

Savings

1500

Other Expenses

1800

The approximate percentages of the monthly budget NOT spent on savings is (A) 10% (B) 14%


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General Aptitudes

(C) 81% EE SP 11.50

Page 605

(D) 86%

A and B are friends. They decide to meet between 1 PM and 2 PM on a given day. There is a conditions that whoever arrives first will not wait for the other for more than 15 minutes. The probability that they will meet on that days is (A) 1/4 (B) 1/16 (C) 7/16 (D) 9/16 2011

EE SP 11.51

There are two candidates P and Q in an election. During the campaign, 40% of voter promised to vote for P , and rest for Q . However, on the day of election 15% of the voters went back on their promise to vote for P and instead voted for Q . 25% of the voter went back on their promise to vote for Q and instead voted for P . Suppose, P lost by 2 votes, then what was the total number of voters ? (A) 100 (B) 110

nodia (C) 90

EE SP 11.52

ONE MARK

(D) 95

The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relations in the original pair : Gladiator : Arena (A) dancer : stage (B) commuter : train (C) teacher : classroom (D) lawyer : courtroom

EE SP 11.53

Choose the most appropriate word from the options given below to complete the following sentence : Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which................... treatments are unsatisfactory. (A) similar (B) most (C) uncommon (D) available

EE SP 11.54

Choose the word from the from the options given below that is most opposite in meaning to the given word : Frequency (A) periodicity (B) rarity (C) gradualness (D) persistency

EE SP 11.55

Choose the most appropriate word from the options given below to complete the following sentence : It was her view that the country’s had been ............. by foreign techno-crafts, so that to invite them to come back would be counter-productive. (A) identified (B) ascertained (C) exacerbated (D) analysed


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General Aptitudes

Chapter 11

2011 EE SP 11.56

TWO MARKS

The fuel consumed by a motor cycle during a journey while travelling at various speed is indicated in the graph below.

nodia

The distance covered during four laps of the journey are listed in the table below Lap

Distance (km)

Average speed (km/hour)

P

15

15

Q

75

45

R

40

75

S 10 10 From the given data, we can conclude that the fuel consumed per kilometre was least during the lap (A) P (B) Q (C) R (D) S

EE SP 11.57

The horse has played a little known but very important role in the field of medicine. Horses were injected with toxins of disease until their blood build up immunities. Then a serum was made from their blood. Serums to fight with diphteria and tetanus were developed this way. It can be inferred from the passage, that horses were (A) given immunity to diseases (B) generally quite immune to diseases (C) given medicines to fight toxins (D) given diphtheria and tetanus serums

EE SP 11.58

The sum of n terms of the series 4 + 44 + 444 + ........ (A) (4/81) [10n + 1 - 9n - 1] (B) (4/81) [10n - 1 - 9n - 1] (C) (4/81) [10n + 1 - 9n - 10] (D) (4/81) [10n - 9n - 10]

EE SP 11.59

Given that f (y) = y /y, and q is any non-zero real number, the value of f (q) - f (- q) is (A) 0 (B) - 1 (C) 1 (D) 2

EE SP 11.60

Three friends R, S and T shared toffee from a bowl. R took 1/3 rd of the toffees, but returned four to the bowl. S took 1/4 th of what was left but returned three


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General Aptitudes

Page 607

toffees to the bowl. T took half of the remainder but returned two back into the bowl. If the bowl had 17 toffees left, how many toffees were originally there in the bowl ? (A) 38 (B) 31 (C) 48

(D) 41

2010 EE SP 11.61

EE SP 11.62

Which of the following options is the closest in meaning to the word below ? Circuitous (A) Cyclic (B) Indirect (C) Confusing (D) Crooked

nodia

The question below consist of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair. Unemployed : Worker (A) Fallow : Land (B) Unaware : Sleeper (C) Wit : Jester

EE SP 11.63

ONE MARK

(D) Renovated : House

Choose the most appropriate word from the options given below to complete the following sentence : If we manage to ........ our natural resources, we would leave a better planet for our children. (A) unhold (B) restrain (C) cherish (D) conserve

EE SP 11.64

Choose the most appropriate word from the options given below to complete the following sentence : His rather casual remarks on politics..................his lack of seriousness about the subject. (A) masked (B) belied (C) betrayed (D) suppressed

EE SP 11.65

25 persons are in a room 15 of them play hockey, 17 of them play football and 10 of them play hockey and football. Then the number of persons playing neither hockey nor football is (A) 2 (B) 17 (C) 13 (D) 3 2010

EE SP 11.66

TWO MARKS

Modern warfare has changed from large scale clashes of armies to suppression of civilian populations. Chemical agents that do their work silently appear to be suited to such warfare ; and regretfully, their exist people in military establishments who think that chemical agents are useful fools for their cause. Which of the following statements best sums up the meaning of the above passage


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General Aptitudes

Chapter 11

? (A) Modern warfare has resulted in civil strife. (B) Chemical agents are useful in modern warfare. (C) Use of chemical agents in ware fare would be undesirable. (D) People in military establishments like to use chemical agents in war. EE SP 11.67

If 137 + 276 = 435 how much is 731 + 672 ? (A) 534 (B) 1403 (C) 1623

EE SP 11.68

(D) 1531

5 skilled workers can build a wall in 20 days; 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall ? (A) 20 days (B) 18 days

nodia (C) 16 days

EE SP 11.69

EE SP 11.70

(D) 15 days

Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how much distinct 4 digit numbers greater than 3000 can be formed ? (A) 50 (B) 51 (C) 52 (D) 54 Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e. brothers and sisters.) All were born on 1st January. The age difference between any two successive siblings (that is born one after another) is less than 3 years. Given the following facts : 1. Hari’s age + Gita’s age > Irfan’s age + Saira’s age. 2. The age difference between Gita and Saira is 1 year. However, Gita is not the oldest and Saira is not the youngest. 3. There are no twins. In what order were they born (oldest first) ? (A) HSIG (B) SGHI (C) IGSH (D) IHSG ***********


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General Aptitudes

Page 609

SOLUTIONS SOL 11.1

Correct option is (B). cope with: presented by To over come any difficulties presented by: adapt to

SOL 11.2

Correct option is (C). By the incomplete statement, we conclude that his performance is just ordinary or not special or medium quality so it comes in mediocre.

SOL 11.3

SOL 11.4

nodia Correct option is (C). He will assume final responsibility. Correct answer is 96. Given that

1 2 bz + z l = 98

...(i)

and we have to determine

1 2 bz + 2 l = ? z

â + b h2 = a2 + b2 + 2a $ b Since, Applying this formula in equation (i), we get 1 2 bz + z l = 98 or or So,

SOL 11.5

z2 + 2.z. 1 + 12 = 98 z z z2 + 12 = 98 - 2 z z2 + 12 = 96 z

Correct option is (D). Given the quadratic equations, ...(i) ax2 + bx + c = 0 2 and ...(ii) ax + b x + c = 0 In equation (ii), x has two values + x and - x . So, it may be resolved into two equations ax2 + bx + c = 0 with roots â 1, a 2h and ax2 - bx + c = 0 with roots ^b 1, b 2h Hence, there are total 4 real roots.

SOL 11.6

Correct option is (C). Passage summary gives the conclusion: The low terrian of the Palghat gap has a significant impact on weather pattern in neighbouring parts of Tamilnadu and Kerala.


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General Aptitudes

Chapter 11

SOL 11.7

Correct option is (B). Strategies are now available for eliminating psychiatric illness.

SOL 11.8

Correct answer is 850. Total round trip fare for group of 5 tourist without discount (i)

(ii)

= 5 # 200 = Rs. 1000 Discount for round trip = 10% of total fare = 10 # 1000 = Rs. 100 100 Discount for having a group of 5 tourist = 5% of total fare = 5 # 1000 = Rs. 50 100

Total discount = Discount for round + Discount for having a group of 5 tourist.

nodia

= 100 + 50 = Rs. 150 Thus, the net round trip fare for group of 5 tourist after discount is Net fare = total fare - total discount = Rs. 1000 - Rs. 150 = Rs. 850

SOL 11.9

Correct answer is 48%. Total respondents = 300 (150 men + 150 women) 150 men = 40 men + 30 men + 60 (men) + 20 (men) (car) (Scooter) (Car & Scooter) (Nothing)

Total number of men who owns car = 100 Total number of men who owns scooter = 90 Total number of men who do not own a scooter = 60 ...(i) = 40 ^carh + 20 ^nothingh 150 women = 34 women + 20 women + 46 (women) + 50 (women) (car) (Scooter) (Car & Scooter) (nothing) Total number of women who owns car = 80 Total number of women who owns scooter = 66 Total number of women who do not own a scooter = 34(Car) + 50 (Nothing) = 84

...(ii)

Percent of respondents who do not own a scooter ^men + womenh who do not own scooter = # 100 Total respondents = 60 + 84 # 100 = 48% 300 SOL 11.10

Correct answer is 6. We draw a tetrahedron structure ABCD as


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General Aptitudes

Page 611

Now, P is a point inside the tetrahedron.

nodia

The point P is connected to each corners A, B, C, D of the tetrahedron. So, we have the internal planes as ABP, APC, BPC, DPC, DPB, DPA i.e the total number of internal planes is 6.

SOL 11.11

Correct option is (A). A pronoun (it) is used after a conjunction (because), so the complete sentence is: India is a post-colonial country because it was a former British colony.

SOL 11.12

Correct option is (B). Who did you say was coming to see us this evening ? This is a question and ‘who’ in that sentence is an interrogative pronoun. For that reason we need to invert subject and verb and write ‘did you say ?’

SOL 11.13

Correct option is (A). Eradicate- to remove or destroy utterly Distort- to give a false, perverted, or disproportionate meaning to; misrepresent Saturate-to soak, impregnate, or imbue thoroughly or completely Utilize-to put to use

SOL 11.14


Average value = 200 # 9 + 100 19


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General Aptitudes

Chapter 11

= 100 SOL 11.15

Correct option is (C). Given expression can be written as x = 12 + 12 + 12 + ... x = 12 + x ax= 2 x = 12 + x x2 - x - 12 = 0 x2 - 4x + 3x - 12 = 0

12 + 12 + ...

x ^x - 4h + 3 ^x - 4h = 0 ^x + 3h^x - 4h = 0

nodia x =- 3 , 4 x =4

SOL 11.16

SOL 11.17

SOL 11.18

(x can not be negative)

Correct option is (A). Ethnic cleansing or makeup is the process of using violent methods to force certain groups of people out of a particular area or country. We can infer that Kaliningrad was historically Russian in its ethnic make up. Correct option is (D). The number of people with malarial fever (also contracted from mosquito bites) has increased this year.

Correct option is (D). x2 - 2x + 3 = 11 So we can have two possible equalities ...(i) x2 - 2x + 3 = 11 or x2 - 2x - 8 = 0 2 2 and ...(ii) x - 2x + 3 =- 11 or x - 2x + 14 = 0 For eq (ii), Discriminant T = b2 - 4ac = 4 - 4 (1) (14) 1 0 So, roots of equation (ii) will be imaginary, but given x is real. Therefore we dont consider this equation. For eq(i) Discriminant T = b2 - 4ac = 4 - 4 (1) (- 8) 1 0 So, roots of equation (i) will be real. x2 - 2x - 8 = 0 x2 - 4x + 2x - 8 = 0 ^x - 4h^x + 2h = 0 x =- 2 , 4 For x =- 2

- x3 + x2 - x = -^- 8h + ^4h - ^- 2h = 8 + 4 + 2 = 14

For x = 4 - x3 + x2 - x

-^64h + ^16h - 4 = 52


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SOL 11.20

General Aptitudes

Page 613

Correct answer is 140. Let male students in 2006 be x1 and female students be y1 . From the given graph we can see that in 2008 ratio of male to female students is 2.5. So x1 = 2.5 y1 x1 = 2.5y1 Let number of male student in 2009 be x2 and number of female students be y2 . From the graph we can see that in 2009, ratio of male to female students is 3. So x2 = 3 y2 Given that number of female students is double in 2009, so we have y2 = 2y1 . The above ratio now becomes as x2 = 3 2y1 x2 = 6y1 % of No. of male students increased in 2009 3 x = x2 - x1 # 100 x1 6y - 2.5y1 3x = 1 100 = 140 2.5y1 #

nodia Correct option is (A).

As shown in Figure above, at 6:00 a.m. initial angle between minute and hour hand is 180c. As we know that hour hand completes 30c in every hour (60 minutes), so angle moved by hour hand in 1 minute is 30 60 . Similarly, minute hand complete 360c in every hour, so angle moved by minute hand in 1 minute is 360 60 . Let us assume that after x minutes the angle is 60c. Let us assume that after x minutes the angle is 60c. 60c = 180c + b 360c l x - b 30c l x 60c 60c angle moved by hour hand

angle moved by min hand

60c = 180c + 6x - 0.5x 5.5x = 120 x = 120c = 21.8 min 6.5 Therefore approximately at 6.22 am the angle will close to 60c between the hour hand and minute hand.


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General Aptitudes

Chapter 11

SOL 11.21

Correct option is (D). IV underlined part is not correct. The correct statement for the IV underline part will be ‘lost consciousness’.

SOL 11.22

Correct option is (C). For a conditional sentence, the rule is as given below. If + past perfect tense Would/could have + past participles. Thus, the complete sentence is If she had known how to calibrate the instrument, she could have done the experiment.

SOL 11.23

Correct option is (C) Coherent – capable of thinking and expressing yourself in a clear and consistent manner. rambling – spreading out in different directions or distributed irregularly. Sticky – covered with an adhesive material. So, coherent and rambling are opposite to each other.

SOL 11.24

SOL 11.25

nodia

Correct option is (C). 2, 5, 10, 17, 26, 37, 50, 64 The difference between successive terms is in AP with initial term 3 and the common difference 2. With this logic 64 is the wrong term and the correct number in place of 64 is 65. Correct option is (C). Total students in class = 44 Total marks scored by class = 21 # 2 + 15 # 3 + 23 # 2 = 42 + 45 + 46 = 133 Average marks = Total marks Total students = 133 44 = 3.02

SOL 11.26

Correct option is (B). (i) Dance program is scheduled for 10 a.m. (ii) Participating student should come at 9 a.m. accompanied by a parent. (iii) non-participating student should come at 10 a.m. with their parents. Combining these three instructions (B) option is most appropriate.

SOL 11.27

Correct option is (D). (i) Clinching evidence was provided by experimental measurement of position of star. (ii) Our understanding of universe changes, so experimental evidence are important.


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General Aptitudes

Page 615

So, by these (iv) is correct option. SOL 11.28

Correct option is (D). Let GDP in 2011-12 be x GDP in 2012-13 is 1.07x (7% grew) During 2012-13 exchange rate increase from Rs. 50/USD to Rs.60/USD. So, GDIP in comparison to USD is obtained as at initial of 2012-13 GDP/USD = x 50 and at final 2012-13 GDP/USD = 1.07x 60 So,

increase and decrease =

1.07x 60

- 50x

x 50

# 100%

nodia = 10.83% = 11%

SOL 11.29

SOL 11.30

Correct option is (C). Let number of female students in 2011 be x . So, Number of male students in 2011 = x (ratio is 1) Number of female students in 2012 = x (Given) Number of male students in 2012 = 1.5x (ratio is 1.5) Ratio of male students in 2012 to male students in 2011 is = 1.5x = 1.5 : 1 x Correct option is (C). Given equation is

^7526h8 - ^Y h8 = ^4364h8

or ^Y h8 = ^7526h8 - ^4364h8 Octal subtraction is done in same way as decimal subtraction. The only difference is that while obtaining carry we get 8 instead of 10. ^7526h8 -^4364h8 ^3142h8 SOL 11.31

Option (D) is correct Two and two make four

SOL 11.32

Correct option is (C). You can always given me a ring whenever you need. Because a friend is need is a friend indeed

SOL 11.33

Correct option is (C). Let the temperature on Monday, Tuesday, Wednesday and Thursday be respectively as TM , TTU , TW , TTH So, from the given data we have


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General Aptitudes

Chapter 11

TH + TTU + TW = 41 3 TTU + TW + TTH = 43 3

and

....(1) ....(2)

also, as the temperature on Thursday was 15% higher than that of Monday ....(3) i.e. TTH = 1.15 TM solving eq (1), (2) and (3), we obtain TTH = 46cC SOL 11.34

SOL 11.35

Correct option is (B). Dare to commit mistakes

nodia Correct option is (D). They were requested not to quarrel with others. Quarrel has a similar meaning to ‘fall out’

SOL 11.36

Correct option is (C). Given, the distance travelled by the car in each quarter intervals as Distance

Time Duration

8 km

1 4

hr

6 km

1 4

hr

16 km

1 4

hr

Therefore, the total time taken = 1 + 1 + 1 + 3 hr 4 4 4 4 Hence,

SOL 11.37

Total distance travelled = 8 + 6 + 16 = 30 km average speed = Total distance travelled Total time taken = 30 = 40 km/hr 3/4

Correct option is (D). It will be easy to check the options for given series. From the given series. 10 + 84 + 734 + ...... We get Sum of 1 term = S1 = 10 Sum of 2 terms = S2 = 10 + 84 = 94 and sum of 3 terms = S 3 = 10 + 84 + 734 = 828 Checking all the options one by one, we observe that only (D) option satisfies as 9 ^9n - 1h + n2 Sn = 8 9 ^92 - 1h so, + 22 = 10 S1 8 9 ^9 - 1h + 22 = 94 S2 = 8 9 ^93 - 1h + 32 = 828 S3 = 8


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General Aptitudes

SOL 11.38

Correct option is (D). Nationalism in India is heterogeneous

SOL 11.39

Correct option is (B). Given, the quadratic equation

Page 617

3x2 + 2x + P ^P - 1h = 0 It will have the roots with opposite sign if

P ^P - 1h < 0 So it can be possible only when

P < 0 and P - 1 > 0 or P > 0 and P - 1 < 0 st The 1 condition tends to no solution for P . Hence, from the second condition, we obtain

nodia 0
SOL 11.40

SOL 11.41

Correct option is (A). In a leap year, there are 366 days So, 52 weeks will have 52 saturdays and for remaining two days ^366 - 52 # 7 = 2h. We can have the following combinations Saturday, Sunday Sunday, Monday Monday, Tuesday Tuesday, Wednesday Wednesday, Thursday Thursday, Friday Friday, Saturday Out of these seven possibilities, only two consist a saturday. Therefore, the probability of saturday is given as P =2 7 Correct option is (D). Let 1.001 = x So in given data : x1259 = 3.52 x2062 = 7.85 x3321 = x1259 + 2062 = x1259 x2062 = 3.52 # 7.85 = 27.64

Again

SOL 11.42


SOL 11.43


SOL 11.44



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General Aptitudes

SOL 11.45


SOL 11.46


SOL 11.47

Correct option is (A). Let no. of notes of Rs.20 be x and no. of notes of Rs. 10 be y . Then from the given data.

Chapter 11

x + y = 14 20x + 10y = 230 Solving the above two equations we get x = 9, y = 5 So, the no. of notes of Rs. 10 is 5.

SOL 11.48

nodia

Correct option is (A). We will categorize the 8 bags in three groups as : (i) A1 A2 A 3 , (ii) B1 B2 B 3 , (iii) C1 C2 Weighting will be done as bellow : 1st weighting " A1 A2 A 3 will be on one side of balance and B1 B2 B 3 on the other. It may have three results as described in the following cases. Case 1 : A1 A 2 A 3 = B 1 B 2 B 3 This results out that either C1 or C2 will heavier for which we will have to perform weighting again. 2 nd weighting " C1 is kept on the one side and C2 on the other. then C1 is heavier. if C1 > C 2 C1 < C 2

then C2 is heavier.

Case 2 : A1 A 2 A 3 > B1 B 2 B 3 it means one of the A1 A2 A 3 will be heavier So we will perform next weighting as: 2 nd weighting " A1 is kept on one side of the balance and A2 on the other. it means A 3 will be heavier if A1 = A 2 then A1 will be heavier A1 > A 2 then A2 will be heavier A1 < A 2 Case 3 : A1 A 2 A 3 < B 1 B 2 B 3 This time one of the B1 B2 B 3 will be heavier, So again as the above case weighting will be done. 2 nd weighting " B1 is kept one side and B2 on the other if B 3 will be heavier B1 = B 2 B1 will be heavier B1 > B 2 B1 < B 2 B2 will be heavier So, as described above, in all the three cases weighting is done only two times to give out the result so minimum no. of weighting required = 2. SOL 11.49

Correct option is (D). Total budget = 4000 + 1200 + 2000 + 1500 + 1800 = 10, 500 The amount spent on saving = 1500 So, the amount not spent on saving = 10, 500 - 1500 = 9000


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General Aptitudes

Page 619

So, percentage of the amount = 9000 # 100% = 86% 10500 SOL 11.50

Option (S) is correct. The graphical representation of their arriving time so that they met is given as below in the figure by shaded region.

nodia So, the area of shaded region is given by Area of 4PQRS

- (Area of TEFQ + Area of

TGSH ) = 60 # 60 - 2 b 1 # 45 # 45 l 2 = 1575 So, the required probability = 1575 = 7 3600 16 SOL 11.51

Correct option is (A). Let us assume total voters are 100. Thus 40 voter (i.e. 40 %) promised to vote for P and 60 (rest 60 % ) promised to vote fore Q. Now, 15% changed from P to Q (15 % out of 40) 15 Changed voter from P to Q 40 = 6 100 # Now Voter for P 40 - 6 = 34 Also, 25% changed form Q to P (out of 60%) 25 Changed voter from Q to P 60 = 15 100 # Now Voter for P 34 + 15 = 49 Thus P P got 49 votes and Q got 51 votes, and P lost by 2 votes, which is given. Therefore 100 voter is true value.

SOL 11.52

Correct option is (A). A gladiator performs in an arena. Commutators use trains. Lawyers performs, but do not entertain like a gladiator. Similarly, teachers educate. Only dancers


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General Aptitudes

Chapter 11

performs on a stage. SOL 11.53

Correct option is (D). Available is appropriate because manipulation of genes will be done when other treatments are not useful.

SOL 11.54

Correct option is (B). Periodicity is almost similar to frequency. Gradualness means something happening with time. Persistency is endurance. Rarity is opposite to frequency.

SOL 11.55

Correct option is (C). The sentence implies that technocrats are counterproductive (negative). Only (C) can bring the same meaning.

SOL 11.56

SOL 11.57

SOL 11.58

nodia

Correct option is (B). Since fuel consumption/litre is asked and not total fuel consumed, only average speed is relevant. Maximum efficiency comes at 45 km/hr, So least fuel consumer per litre in lap Q Correct option is (B). Option B fits the sentence, as they built up immunities which helped humans create serums from their blood. Correct option is (C). 4 + 44 + 444 + .............. 4 (1 + 11 + 111 + .......) = 4 (9 + 99 + 999 + ............) 9 = 4 [(10 - 1) + (100 - 1) + ........] 9 = 4 [10 (1 + 10 + 102 + 103) - n] 9 n = 4 :10 # 10 - 1 - nD 9 10 - 1 = 4 610n + 1 - 10 - 9n@ 81

SOL 11.59

Correct option is (D). y y -y f (- y) = =- f (y) y f (q) - f (- q) = 2f (q) = 2 f (y) =

Now or SOL 11.60

Correct option is (C). Let total no of toffees be x . The following table shows the all calculations. Friend

Bowl Status


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General Aptitudes

Page 621

R

= x -4 3

= 2x + 4 3

S

= 1 :2x + 4D - 3 4 3 = x +1-3 = x -2 6 6

= 2x + 4 - x + 2 3 6 = x +6 2

T

= 1 a x + 6k - 2 2 2 = x +1 4

= x +6-x -1 2 4 = x +5 4

x + 5 = 17 4 x = 17 - 5 = 12 4

Now,

nodia or

x = 12 # 4 = 48

SOL 11.61

Correct option is (B). Circuitous means round about or not direct. Indirect is closest in meaning to this circuitous (A) Cyclic : Recurring in nature (B) Indirect (C) Confusing (D) Crooked

: Not direct : lacking clarity of meaning : set at an angle; not straight

SOL 11.62

Correct option is (B). A worker may by unemployed. Like in same relation a sleeper may be unaware.

SOL 11.63

Correct option is (D). Here conserve is most appropriate word.

SOL 11.64

Correct option is (C). Betrayed means reveal unintentionally that is most appropriate.

SOL 11.65

Correct option is (D). Number of people who play hockey n (A) = 15 Number of people who play football n (B) = 17 Persons who play both hockey and football n (A + B) = 10 Persons who play either hockey or football or both : n (A , B) = n (A) + n (B) - n (A + B) = 15 + 17 - 10 = 22 Thus people who play neither hockey nor football = 25 - 22 = 3

SOL 11.66


SOL 11.67

Correct option is (C). Since 7 + 6 = 13 but unit digit is 5 so base may be 8 as 5 is the remainder when


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General Aptitudes

13 is divided by 8. Let us check. 137 8 276 8 435 SOL 11.68

Thus here base is 8. Now

Chapter 11

731 8 672 8 1623

Correct option is (D). Let W be the total work. =W 20 Per day work of one skill worker = W =W 5 # 20 100 Similarly per day work of 1 semi-skilled workers = W = W 8 # 25 200 Similarly per day work of one semi-skill worker = W = W 10 # 30 300 Thus total per day work of 2 skilled, 6 semi-skilled and 5 unskilled workers is = 2W + 6W + 5W = 12W + 18W + 10W = W 100 200 300 600 15 Therefore time to complete the work is 15 days. Per day work of 5 skilled workers

SOL 11.69

nodia

Correct option is (B). As the number must be greater than 3000, it must be start with 3 or 4. Thus we have two case: Case (1) If left most digit is 3 an other three digits are any of 2, 2, 3, 3, 4, 4, 4, 4. (1) Using 2, 2, 3 we have 3223, 3232, 3322 i.e. 3! = 3 no. 2! (2) Using 2, 2, 4 we have 3224, 3242, 3422 i.e. 3! = 3 no. 2! (3) Using 2, 3, 3 we have 3233, 3323, 3332 i.e. 3! = 3 no. 2!

(4) Using 2, 3, 4 we have 3! = 6 no. (5) Using 2, 4, 4 we have 3244, 3424, 3442 i.e. 3! = 3 no. 2! (6) Using 3, 3, 4 we have 3334, 3343, 3433 i.e. 3! = 3 no. 2! (7) Using 3, 4, 4 we have 3344, 3434, 3443 i.e. 3! = 3 no. 2! (8) Using 4, 4, 4 we have 3444 i.e. 3! = 1 no. 3! Total 4 digit numbers in this case is 1 + 3 + 3 + 3 + 6 + 3 + 3 + 3 + 1 = 25 Case 2 : If left most is 4 and other three digits are any of 2, 2, 3, 3, 3, 4, 4, 4. (1) Using 2, 2, 3 we have 4223, 4232, 4322 i.e. . 3! = 3 no 2! (2) Using 2, 2, 4 we have 4224, 4242, 4422 i.e. . 3! = 3 no 2! (3) Using 2, 3, 3 we have 4233, 4323, 4332 i.e. . 3! = 3 no 2! (4) Using 2, 3, 4 we have i.e. . 3! = 6 no

(5) Using 2, 4, 4 we have 4244, 4424, 4442 i.e. . 3! = 3 no 2!


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General Aptitudes

Page 623

(6) Using 3, 3, 3 we have 4333 i.e 3! = 1. no. 3! (7) Using 3, 3, 4 we have 4334, 4343, 4433 i.e. . 3! = 3 no 2! (8) Using 3, 4, 4 we have 4344, 4434, 4443 i.e. . 3! = 3 no 2! (9) Using 4, 4, 4 we have 4444 i.e. 3! = 1. no 3! Total 4 digit numbers in 2nd case = 3 + 3 + 3 + 6 + 3 + 3 + 1 + 3 + 1 = 26 Thus total 4 digit numbers using case (1) and case (2) is = 25 + 26 = 51 SOL 11.70

Correct option is (B). Let H , G , S and I be ages of Hari, Gita, Saira and Irfan respectively.

nodia

Now from statement (1) we have H + G > I + S Form statement (2) we get that G - S = 1 or S - G = 1 As G can’t be oldest and S can’t be youngest thus either GS or SG possible. From statement (3) we get that there are no twins (A) HSIG : There is I between S and G which is not possible (B) SGHI : SG order is also here and S > G > H > I and G + H > S + I which is possible. (C) IGSH : This gives I > G and S > H and adding these both inequalities we have I + S > H + G which is not possible. (D) IHSG : This gives I > H and S > G and adding these both inequalities we have I + S > H + G which is not possible. **********


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